Prove $sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta $
$begingroup$
Prove $$sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$$
I only know how to solve using factoring and the basic trig identities, I do not know reduction or anything of the sort, please prove using the basic trigonometric identities and factoring.
After some help I found that you move the identity around, so:
$sin^2(theta)-cos^2(theta)=sin^4(theta)-cos^4(theta)$
Then,
$sin^2(theta)-cos^2(theta)=(sin^2(theta)+cos^2(theta))(sin^2(theta)-cos^2(theta))$
the positive sum of squares defaults to 1 and then the right side equals the left, but how does that prove the original identity?
trigonometry
edited Nov 11 '17 at 5:21
Paramanand Singh
51.1k557168
asked Aug 29 '15 at 23:30
Sunny MannSunny Mann
36118
$endgroup$
add a comment |
$begingroup$
Prove $$sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$$
I only know how to solve using factoring and the basic trig identities, I do not know reduction or anything of the sort, please prove using the basic trigonometric identities and factoring.
After some help I found that you move the identity around, so:
$sin^2(theta)-cos^2(theta)=sin^4(theta)-cos^4(theta)$
Then,
$sin^2(theta)-cos^2(theta)=(sin^2(theta)+cos^2(theta))(sin^2(theta)-cos^2(theta))$
the positive sum of squares defaults to 1 and then the right side equals the left, but how does that prove the original identity?
trigonometry
edited Nov 11 '17 at 5:21
Paramanand Singh
51.1k557168
asked Aug 29 '15 at 23:30
Sunny MannSunny Mann
36118
$endgroup$
$begingroup$
Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
$endgroup$
– John Joy
Aug 30 '15 at 0:23
add a comment |
$begingroup$
Prove $$sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$$
I only know how to solve using factoring and the basic trig identities, I do not know reduction or anything of the sort, please prove using the basic trigonometric identities and factoring.
After some help I found that you move the identity around, so:
$sin^2(theta)-cos^2(theta)=sin^4(theta)-cos^4(theta)$
Then,
$sin^2(theta)-cos^2(theta)=(sin^2(theta)+cos^2(theta))(sin^2(theta)-cos^2(theta))$
the positive sum of squares defaults to 1 and then the right side equals the left, but how does that prove the original identity?
trigonometry
edited Nov 11 '17 at 5:21
Paramanand Singh
51.1k557168
asked Aug 29 '15 at 23:30
Sunny MannSunny Mann
36118
$endgroup$
Prove $$sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$$
I only know how to solve using factoring and the basic trig identities, I do not know reduction or anything of the sort, please prove using the basic trigonometric identities and factoring.
After some help I found that you move the identity around, so:
$sin^2(theta)-cos^2(theta)=sin^4(theta)-cos^4(theta)$
Then,
$sin^2(theta)-cos^2(theta)=(sin^2(theta)+cos^2(theta))(sin^2(theta)-cos^2(theta))$
the positive sum of squares defaults to 1 and then the right side equals the left, but how does that prove the original identity?
trigonometry
trigonometry
edited Nov 11 '17 at 5:21
Paramanand Singh
51.1k557168
asked Aug 29 '15 at 23:30
Sunny MannSunny Mann
36118
edited Nov 11 '17 at 5:21
Paramanand Singh
51.1k557168
asked Aug 29 '15 at 23:30
Sunny MannSunny Mann
36118
edited Nov 11 '17 at 5:21
Paramanand Singh
51.1k557168
edited Nov 11 '17 at 5:21
Paramanand Singh
51.1k557168
edited Nov 11 '17 at 5:21
Paramanand Singh
51.1k557168
51.1k557168
asked Aug 29 '15 at 23:30
Sunny MannSunny Mann
36118
asked Aug 29 '15 at 23:30
Sunny MannSunny Mann
36118
asked Aug 29 '15 at 23:30
Sunny MannSunny Mann
36118
36118
$begingroup$
Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
$endgroup$
– John Joy
Aug 30 '15 at 0:23
add a comment |
$begingroup$
Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
$endgroup$
– John Joy
Aug 30 '15 at 0:23
$begingroup$
Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
$endgroup$
– John Joy
Aug 30 '15 at 0:23
$begingroup$
Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
$endgroup$
– John Joy
Aug 30 '15 at 0:23
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $sin^2theta + cos^2theta = 1 $ rearranged into $sin^2theta = 1 - cos^2theta $ and $cos^2theta = 1 - sin^2theta $
We can see that: $cos^4theta = cos^2thetacos^2theta = (1-sin^2theta)(1-sin^2theta) = 1-2sin^2theta + sin^4theta $
$sin^2theta + cos^4theta = cos^2theta + sin^4theta $
$sin^2theta + (1-2sin^2theta + sin^4theta) = cos^2theta + sin^4theta $
$sin^2theta + 1-2sin^2theta + sin^4theta = cos^2theta + sin^4theta $
$sin^4theta-sin^2theta+1= cos^2theta + sin^4theta $
$sin^4theta-(1-cos^2theta)+1=cos^2theta + sin^4theta $
$sin^4theta-1+cos^2theta+1=cos^2theta + sin^4theta $
$sin^4theta+cos^2theta=cos^2theta + sin^4theta $
answered Aug 30 '15 at 0:22
LeoLeo
908
$endgroup$
$begingroup$
(1−2sin2θ+sin4θ) where did you get this expression from?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:30
1
$begingroup$
Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
$endgroup$
– Leo
Aug 30 '15 at 0:30
$begingroup$
you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:34
$begingroup$
We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
$endgroup$
– Leo
Aug 30 '15 at 0:37
$begingroup$
it's supposed to be $1- sin^2(theta), no?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:40
|
show 2 more comments
$begingroup$
Rewrite this as
$$
sin^2 theta - cos^2 theta = sin^4 theta - cos^4 theta
$$
and then factor the right-hand side as a difference of two squares.
answered Aug 29 '15 at 23:32
MicahMicah
30.2k1364106
$endgroup$
$begingroup$
How does this prove the identity?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:34
2
$begingroup$
Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
$endgroup$
– Micah
Aug 29 '15 at 23:34
$begingroup$
could you factor any side actually both of them look like difference of squares?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:39
$begingroup$
You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
$endgroup$
– Micah
Aug 29 '15 at 23:40
$begingroup$
Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:44
|
show 2 more comments
$begingroup$
As $cos^2 theta +sin^2 theta= 1$ we have $$cos^4 theta -sin^4 theta =(cos^2 theta -sin^2 theta)color{red}{(cos^2 theta +sin^2 theta)}= cos^2 theta -sin^2 theta$$
That is,
$$sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta$$
edited Jan 14 at 23:10
J. W. Tanner
3,9071320
answered Nov 10 '17 at 15:59
Guy FsoneGuy Fsone
17.3k43074
$endgroup$
add a comment |
$begingroup$
Here would be the other points to remember:
$sin^2theta+cos^2theta=1$
$x^4-y^4=(x^2-y^2)(x^2+y^2)$
answered Aug 29 '15 at 23:40
JB KingJB King
3,50911014
$endgroup$
add a comment |
$begingroup$
Here's an alternative. I'm not quite sure if this qualifies as a proof, but I think it's an interesting fact:
Consider the function
$$f(theta) = sin^2{theta} - cos^2{theta} - sin^4 theta + cos^4 theta $$
Thus, $f'(theta) = 4 sintheta costheta , ( 1 - sin^2theta - cos^2 theta ) = 0$ and $f$ is therefore constant for any $theta$. We discover this constant is 0 since $f(0) = 0$.
Hope you find this useful/interesting.
answered Oct 11 '17 at 19:50
DmorenoDmoreno
6,53231240
$endgroup$
add a comment |
$begingroup$
A "forwards" proof:
Render
$sin^2theta+cos^4theta=sin^2theta+(1-cos^2theta)^2=sin^2theta+(1-2sin^2theta+sin^4theta)$
Regroup the terms on the right as
$(sin^2theta+1-2sin^2theta)+sin^4theta$
and put $1-sin^2theta=cos^2theta$.
answered Nov 10 '17 at 16:33
Oscar LanziOscar Lanzi
13.3k12136
$endgroup$
add a comment |
$begingroup$
$$sin^2theta+cos^4theta=sin^2theta+bigg(cos^2thetabigg)^2=sin^2theta+bigg(1-sin^2thetabigg)^2=dots$$
answered Aug 30 '15 at 0:13
John JoyJohn Joy
6,29911727
$endgroup$
$begingroup$
could you elaborate please?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:14
$begingroup$
Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
$endgroup$
– John Joy
Aug 30 '15 at 0:20
$begingroup$
$sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:26
$begingroup$
What do you get when you expand $(1-sin^2theta)^2$?
$endgroup$
– John Joy
Aug 30 '15 at 0:28
$begingroup$
$1- sin^4(theta) or cos^4(theta)$
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:36
|
show 9 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1414134%2fprove-sin2-theta-cos4-theta-cos2-theta-sin4-theta%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $sin^2theta + cos^2theta = 1 $ rearranged into $sin^2theta = 1 - cos^2theta $ and $cos^2theta = 1 - sin^2theta $
We can see that: $cos^4theta = cos^2thetacos^2theta = (1-sin^2theta)(1-sin^2theta) = 1-2sin^2theta + sin^4theta $
$sin^2theta + cos^4theta = cos^2theta + sin^4theta $
$sin^2theta + (1-2sin^2theta + sin^4theta) = cos^2theta + sin^4theta $
$sin^2theta + 1-2sin^2theta + sin^4theta = cos^2theta + sin^4theta $
$sin^4theta-sin^2theta+1= cos^2theta + sin^4theta $
$sin^4theta-(1-cos^2theta)+1=cos^2theta + sin^4theta $
$sin^4theta-1+cos^2theta+1=cos^2theta + sin^4theta $
$sin^4theta+cos^2theta=cos^2theta + sin^4theta $
answered Aug 30 '15 at 0:22
LeoLeo
908
$endgroup$
$begingroup$
(1−2sin2θ+sin4θ) where did you get this expression from?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:30
1
$begingroup$
Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
$endgroup$
– Leo
Aug 30 '15 at 0:30
$begingroup$
you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:34
$begingroup$
We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
$endgroup$
– Leo
Aug 30 '15 at 0:37
$begingroup$
it's supposed to be $1- sin^2(theta), no?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:40
|
show 2 more comments
$begingroup$
I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $sin^2theta + cos^2theta = 1 $ rearranged into $sin^2theta = 1 - cos^2theta $ and $cos^2theta = 1 - sin^2theta $
We can see that: $cos^4theta = cos^2thetacos^2theta = (1-sin^2theta)(1-sin^2theta) = 1-2sin^2theta + sin^4theta $
$sin^2theta + cos^4theta = cos^2theta + sin^4theta $
$sin^2theta + (1-2sin^2theta + sin^4theta) = cos^2theta + sin^4theta $
$sin^2theta + 1-2sin^2theta + sin^4theta = cos^2theta + sin^4theta $
$sin^4theta-sin^2theta+1= cos^2theta + sin^4theta $
$sin^4theta-(1-cos^2theta)+1=cos^2theta + sin^4theta $
$sin^4theta-1+cos^2theta+1=cos^2theta + sin^4theta $
$sin^4theta+cos^2theta=cos^2theta + sin^4theta $
answered Aug 30 '15 at 0:22
LeoLeo
908
$endgroup$
$begingroup$
(1−2sin2θ+sin4θ) where did you get this expression from?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:30
1
$begingroup$
Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
$endgroup$
– Leo
Aug 30 '15 at 0:30
$begingroup$
you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:34
$begingroup$
We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
$endgroup$
– Leo
Aug 30 '15 at 0:37
$begingroup$
it's supposed to be $1- sin^2(theta), no?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:40
|
show 2 more comments
$begingroup$
I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $sin^2theta + cos^2theta = 1 $ rearranged into $sin^2theta = 1 - cos^2theta $ and $cos^2theta = 1 - sin^2theta $
We can see that: $cos^4theta = cos^2thetacos^2theta = (1-sin^2theta)(1-sin^2theta) = 1-2sin^2theta + sin^4theta $
$sin^2theta + cos^4theta = cos^2theta + sin^4theta $
$sin^2theta + (1-2sin^2theta + sin^4theta) = cos^2theta + sin^4theta $
$sin^2theta + 1-2sin^2theta + sin^4theta = cos^2theta + sin^4theta $
$sin^4theta-sin^2theta+1= cos^2theta + sin^4theta $
$sin^4theta-(1-cos^2theta)+1=cos^2theta + sin^4theta $
$sin^4theta-1+cos^2theta+1=cos^2theta + sin^4theta $
$sin^4theta+cos^2theta=cos^2theta + sin^4theta $
answered Aug 30 '15 at 0:22
LeoLeo
908
$endgroup$
I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: $sin^2theta + cos^2theta = 1 $ rearranged into $sin^2theta = 1 - cos^2theta $ and $cos^2theta = 1 - sin^2theta $
We can see that: $cos^4theta = cos^2thetacos^2theta = (1-sin^2theta)(1-sin^2theta) = 1-2sin^2theta + sin^4theta $
$sin^2theta + cos^4theta = cos^2theta + sin^4theta $
$sin^2theta + (1-2sin^2theta + sin^4theta) = cos^2theta + sin^4theta $
$sin^2theta + 1-2sin^2theta + sin^4theta = cos^2theta + sin^4theta $
$sin^4theta-sin^2theta+1= cos^2theta + sin^4theta $
$sin^4theta-(1-cos^2theta)+1=cos^2theta + sin^4theta $
$sin^4theta-1+cos^2theta+1=cos^2theta + sin^4theta $
$sin^4theta+cos^2theta=cos^2theta + sin^4theta $
answered Aug 30 '15 at 0:22
LeoLeo
908
answered Aug 30 '15 at 0:22
LeoLeo
908
answered Aug 30 '15 at 0:22
LeoLeo
908
answered Aug 30 '15 at 0:22
LeoLeo
908
908
$begingroup$
(1−2sin2θ+sin4θ) where did you get this expression from?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:30
1
$begingroup$
Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
$endgroup$
– Leo
Aug 30 '15 at 0:30
$begingroup$
you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:34
$begingroup$
We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
$endgroup$
– Leo
Aug 30 '15 at 0:37
$begingroup$
it's supposed to be $1- sin^2(theta), no?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:40
|
show 2 more comments
$begingroup$
(1−2sin2θ+sin4θ) where did you get this expression from?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:30
1
$begingroup$
Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
$endgroup$
– Leo
Aug 30 '15 at 0:30
$begingroup$
you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:34
$begingroup$
We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
$endgroup$
– Leo
Aug 30 '15 at 0:37
$begingroup$
it's supposed to be $1- sin^2(theta), no?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:40
$begingroup$
(1−2sin2θ+sin4θ) where did you get this expression from?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:30
$begingroup$
(1−2sin2θ+sin4θ) where did you get this expression from?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:30
1
1
$begingroup$
Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
$endgroup$
– Leo
Aug 30 '15 at 0:30
$begingroup$
Expanding Brackets. Imagine the sin2(θ) as (sin(θ))^2. Then (1−sin2θ)(1−sin2θ) = (1)(1) + (1)(-sin2θ) + (1)(-sin2θ) + (-sin2θ)(-sin2θ) = 1 - sin2θ - sin2θ + sin4θ ) = 1 - 2sinθ + sin4θ
$endgroup$
– Leo
Aug 30 '15 at 0:30
$begingroup$
you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:34
$begingroup$
you jumped from +$cos^4(theta)$ to ($1-2sin^2(theta)+sin^4(theta))$ How?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:34
$begingroup$
We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
$endgroup$
– Leo
Aug 30 '15 at 0:37
$begingroup$
We can write $cos^4theta $ as $cos^2thetacos^2theta $. Imagine it as $(costheta)^4 = (costheta)^2(costheta)^2 $.
$endgroup$
– Leo
Aug 30 '15 at 0:37
$begingroup$
it's supposed to be $1- sin^2(theta), no?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:40
$begingroup$
it's supposed to be $1- sin^2(theta), no?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:40
|
show 2 more comments
$begingroup$
Rewrite this as
$$
sin^2 theta - cos^2 theta = sin^4 theta - cos^4 theta
$$
and then factor the right-hand side as a difference of two squares.
answered Aug 29 '15 at 23:32
MicahMicah
30.2k1364106
$endgroup$
$begingroup$
How does this prove the identity?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:34
2
$begingroup$
Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
$endgroup$
– Micah
Aug 29 '15 at 23:34
$begingroup$
could you factor any side actually both of them look like difference of squares?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:39
$begingroup$
You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
$endgroup$
– Micah
Aug 29 '15 at 23:40
$begingroup$
Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:44
|
show 2 more comments
$begingroup$
Rewrite this as
$$
sin^2 theta - cos^2 theta = sin^4 theta - cos^4 theta
$$
and then factor the right-hand side as a difference of two squares.
answered Aug 29 '15 at 23:32
MicahMicah
30.2k1364106
$endgroup$
$begingroup$
How does this prove the identity?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:34
2
$begingroup$
Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
$endgroup$
– Micah
Aug 29 '15 at 23:34
$begingroup$
could you factor any side actually both of them look like difference of squares?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:39
$begingroup$
You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
$endgroup$
– Micah
Aug 29 '15 at 23:40
$begingroup$
Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:44
|
show 2 more comments
$begingroup$
Rewrite this as
$$
sin^2 theta - cos^2 theta = sin^4 theta - cos^4 theta
$$
and then factor the right-hand side as a difference of two squares.
answered Aug 29 '15 at 23:32
MicahMicah
30.2k1364106
$endgroup$
Rewrite this as
$$
sin^2 theta - cos^2 theta = sin^4 theta - cos^4 theta
$$
and then factor the right-hand side as a difference of two squares.
answered Aug 29 '15 at 23:32
MicahMicah
30.2k1364106
answered Aug 29 '15 at 23:32
MicahMicah
30.2k1364106
answered Aug 29 '15 at 23:32
MicahMicah
30.2k1364106
answered Aug 29 '15 at 23:32
MicahMicah
30.2k1364106
30.2k1364106
$begingroup$
How does this prove the identity?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:34
2
$begingroup$
Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
$endgroup$
– Micah
Aug 29 '15 at 23:34
$begingroup$
could you factor any side actually both of them look like difference of squares?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:39
$begingroup$
You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
$endgroup$
– Micah
Aug 29 '15 at 23:40
$begingroup$
Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:44
|
show 2 more comments
$begingroup$
How does this prove the identity?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:34
2
$begingroup$
Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
$endgroup$
– Micah
Aug 29 '15 at 23:34
$begingroup$
could you factor any side actually both of them look like difference of squares?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:39
$begingroup$
You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
$endgroup$
– Micah
Aug 29 '15 at 23:40
$begingroup$
Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:44
$begingroup$
How does this prove the identity?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:34
$begingroup$
How does this prove the identity?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:34
2
2
$begingroup$
Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
$endgroup$
– Micah
Aug 29 '15 at 23:34
$begingroup$
Try it and see! What do you get when you factor the right-hand side as a difference of two squares?
$endgroup$
– Micah
Aug 29 '15 at 23:34
$begingroup$
could you factor any side actually both of them look like difference of squares?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:39
$begingroup$
could you factor any side actually both of them look like difference of squares?
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:39
$begingroup$
You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
$endgroup$
– Micah
Aug 29 '15 at 23:40
$begingroup$
You could factor either side, but factoring the right side will be helpful and factoring the left side will not.
$endgroup$
– Micah
Aug 29 '15 at 23:40
$begingroup$
Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:44
$begingroup$
Ok so I get (sin^2(theta)-cos^2(theta))(sin^2(theta)+cos^2(theta)).
$endgroup$
– Sunny Mann
Aug 29 '15 at 23:44
|
show 2 more comments
$begingroup$
As $cos^2 theta +sin^2 theta= 1$ we have $$cos^4 theta -sin^4 theta =(cos^2 theta -sin^2 theta)color{red}{(cos^2 theta +sin^2 theta)}= cos^2 theta -sin^2 theta$$
That is,
$$sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta$$
edited Jan 14 at 23:10
J. W. Tanner
3,9071320
answered Nov 10 '17 at 15:59
Guy FsoneGuy Fsone
17.3k43074
$endgroup$
add a comment |
$begingroup$
As $cos^2 theta +sin^2 theta= 1$ we have $$cos^4 theta -sin^4 theta =(cos^2 theta -sin^2 theta)color{red}{(cos^2 theta +sin^2 theta)}= cos^2 theta -sin^2 theta$$
That is,
$$sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta$$
edited Jan 14 at 23:10
J. W. Tanner
3,9071320
answered Nov 10 '17 at 15:59
Guy FsoneGuy Fsone
17.3k43074
$endgroup$
add a comment |
$begingroup$
As $cos^2 theta +sin^2 theta= 1$ we have $$cos^4 theta -sin^4 theta =(cos^2 theta -sin^2 theta)color{red}{(cos^2 theta +sin^2 theta)}= cos^2 theta -sin^2 theta$$
That is,
$$sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta$$
edited Jan 14 at 23:10
J. W. Tanner
3,9071320
answered Nov 10 '17 at 15:59
Guy FsoneGuy Fsone
17.3k43074
$endgroup$
As $cos^2 theta +sin^2 theta= 1$ we have $$cos^4 theta -sin^4 theta =(cos^2 theta -sin^2 theta)color{red}{(cos^2 theta +sin^2 theta)}= cos^2 theta -sin^2 theta$$
That is,
$$sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta$$
edited Jan 14 at 23:10
J. W. Tanner
3,9071320
answered Nov 10 '17 at 15:59
Guy FsoneGuy Fsone
17.3k43074
edited Jan 14 at 23:10
J. W. Tanner
3,9071320
edited Jan 14 at 23:10
J. W. Tanner
3,9071320
edited Jan 14 at 23:10
J. W. Tanner
3,9071320
3,9071320
answered Nov 10 '17 at 15:59
Guy FsoneGuy Fsone
17.3k43074
answered Nov 10 '17 at 15:59
Guy FsoneGuy Fsone
17.3k43074
answered Nov 10 '17 at 15:59
Guy FsoneGuy Fsone
17.3k43074
17.3k43074
add a comment |
add a comment |
$begingroup$
Here would be the other points to remember:
$sin^2theta+cos^2theta=1$
$x^4-y^4=(x^2-y^2)(x^2+y^2)$
answered Aug 29 '15 at 23:40
JB KingJB King
3,50911014
$endgroup$
add a comment |
$begingroup$
Here would be the other points to remember:
$sin^2theta+cos^2theta=1$
$x^4-y^4=(x^2-y^2)(x^2+y^2)$
answered Aug 29 '15 at 23:40
JB KingJB King
3,50911014
$endgroup$
add a comment |
$begingroup$
Here would be the other points to remember:
$sin^2theta+cos^2theta=1$
$x^4-y^4=(x^2-y^2)(x^2+y^2)$
answered Aug 29 '15 at 23:40
JB KingJB King
3,50911014
$endgroup$
Here would be the other points to remember:
$sin^2theta+cos^2theta=1$
$x^4-y^4=(x^2-y^2)(x^2+y^2)$
answered Aug 29 '15 at 23:40
JB KingJB King
3,50911014
answered Aug 29 '15 at 23:40
JB KingJB King
3,50911014
answered Aug 29 '15 at 23:40
JB KingJB King
3,50911014
answered Aug 29 '15 at 23:40
JB KingJB King
3,50911014
3,50911014
add a comment |
add a comment |
$begingroup$
Here's an alternative. I'm not quite sure if this qualifies as a proof, but I think it's an interesting fact:
Consider the function
$$f(theta) = sin^2{theta} - cos^2{theta} - sin^4 theta + cos^4 theta $$
Thus, $f'(theta) = 4 sintheta costheta , ( 1 - sin^2theta - cos^2 theta ) = 0$ and $f$ is therefore constant for any $theta$. We discover this constant is 0 since $f(0) = 0$.
Hope you find this useful/interesting.
answered Oct 11 '17 at 19:50
DmorenoDmoreno
6,53231240
$endgroup$
add a comment |
$begingroup$
Here's an alternative. I'm not quite sure if this qualifies as a proof, but I think it's an interesting fact:
Consider the function
$$f(theta) = sin^2{theta} - cos^2{theta} - sin^4 theta + cos^4 theta $$
Thus, $f'(theta) = 4 sintheta costheta , ( 1 - sin^2theta - cos^2 theta ) = 0$ and $f$ is therefore constant for any $theta$. We discover this constant is 0 since $f(0) = 0$.
Hope you find this useful/interesting.
answered Oct 11 '17 at 19:50
DmorenoDmoreno
6,53231240
$endgroup$
add a comment |
$begingroup$
Here's an alternative. I'm not quite sure if this qualifies as a proof, but I think it's an interesting fact:
Consider the function
$$f(theta) = sin^2{theta} - cos^2{theta} - sin^4 theta + cos^4 theta $$
Thus, $f'(theta) = 4 sintheta costheta , ( 1 - sin^2theta - cos^2 theta ) = 0$ and $f$ is therefore constant for any $theta$. We discover this constant is 0 since $f(0) = 0$.
Hope you find this useful/interesting.
answered Oct 11 '17 at 19:50
DmorenoDmoreno
6,53231240
$endgroup$
Here's an alternative. I'm not quite sure if this qualifies as a proof, but I think it's an interesting fact:
Consider the function
$$f(theta) = sin^2{theta} - cos^2{theta} - sin^4 theta + cos^4 theta $$
Thus, $f'(theta) = 4 sintheta costheta , ( 1 - sin^2theta - cos^2 theta ) = 0$ and $f$ is therefore constant for any $theta$. We discover this constant is 0 since $f(0) = 0$.
Hope you find this useful/interesting.
answered Oct 11 '17 at 19:50
DmorenoDmoreno
6,53231240
answered Oct 11 '17 at 19:50
DmorenoDmoreno
6,53231240
answered Oct 11 '17 at 19:50
DmorenoDmoreno
6,53231240
answered Oct 11 '17 at 19:50
DmorenoDmoreno
6,53231240
6,53231240
add a comment |
add a comment |
$begingroup$
A "forwards" proof:
Render
$sin^2theta+cos^4theta=sin^2theta+(1-cos^2theta)^2=sin^2theta+(1-2sin^2theta+sin^4theta)$
Regroup the terms on the right as
$(sin^2theta+1-2sin^2theta)+sin^4theta$
and put $1-sin^2theta=cos^2theta$.
answered Nov 10 '17 at 16:33
Oscar LanziOscar Lanzi
13.3k12136
$endgroup$
add a comment |
$begingroup$
A "forwards" proof:
Render
$sin^2theta+cos^4theta=sin^2theta+(1-cos^2theta)^2=sin^2theta+(1-2sin^2theta+sin^4theta)$
Regroup the terms on the right as
$(sin^2theta+1-2sin^2theta)+sin^4theta$
and put $1-sin^2theta=cos^2theta$.
answered Nov 10 '17 at 16:33
Oscar LanziOscar Lanzi
13.3k12136
$endgroup$
add a comment |
$begingroup$
A "forwards" proof:
Render
$sin^2theta+cos^4theta=sin^2theta+(1-cos^2theta)^2=sin^2theta+(1-2sin^2theta+sin^4theta)$
Regroup the terms on the right as
$(sin^2theta+1-2sin^2theta)+sin^4theta$
and put $1-sin^2theta=cos^2theta$.
answered Nov 10 '17 at 16:33
Oscar LanziOscar Lanzi
13.3k12136
$endgroup$
A "forwards" proof:
Render
$sin^2theta+cos^4theta=sin^2theta+(1-cos^2theta)^2=sin^2theta+(1-2sin^2theta+sin^4theta)$
Regroup the terms on the right as
$(sin^2theta+1-2sin^2theta)+sin^4theta$
and put $1-sin^2theta=cos^2theta$.
answered Nov 10 '17 at 16:33
Oscar LanziOscar Lanzi
13.3k12136
edited Nov 10 '17 at 16:40
answered Nov 10 '17 at 16:33
Oscar LanziOscar Lanzi
13.3k12136
answered Nov 10 '17 at 16:33
Oscar LanziOscar Lanzi
13.3k12136
answered Nov 10 '17 at 16:33
Oscar LanziOscar Lanzi
13.3k12136
13.3k12136
add a comment |
add a comment |
$begingroup$
$$sin^2theta+cos^4theta=sin^2theta+bigg(cos^2thetabigg)^2=sin^2theta+bigg(1-sin^2thetabigg)^2=dots$$
answered Aug 30 '15 at 0:13
John JoyJohn Joy
6,29911727
$endgroup$
$begingroup$
could you elaborate please?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:14
$begingroup$
Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
$endgroup$
– John Joy
Aug 30 '15 at 0:20
$begingroup$
$sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:26
$begingroup$
What do you get when you expand $(1-sin^2theta)^2$?
$endgroup$
– John Joy
Aug 30 '15 at 0:28
$begingroup$
$1- sin^4(theta) or cos^4(theta)$
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:36
|
show 9 more comments
$begingroup$
$$sin^2theta+cos^4theta=sin^2theta+bigg(cos^2thetabigg)^2=sin^2theta+bigg(1-sin^2thetabigg)^2=dots$$
answered Aug 30 '15 at 0:13
John JoyJohn Joy
6,29911727
$endgroup$
$begingroup$
could you elaborate please?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:14
$begingroup$
Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
$endgroup$
– John Joy
Aug 30 '15 at 0:20
$begingroup$
$sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:26
$begingroup$
What do you get when you expand $(1-sin^2theta)^2$?
$endgroup$
– John Joy
Aug 30 '15 at 0:28
$begingroup$
$1- sin^4(theta) or cos^4(theta)$
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:36
|
show 9 more comments
$begingroup$
$$sin^2theta+cos^4theta=sin^2theta+bigg(cos^2thetabigg)^2=sin^2theta+bigg(1-sin^2thetabigg)^2=dots$$
answered Aug 30 '15 at 0:13
John JoyJohn Joy
6,29911727
$endgroup$
$$sin^2theta+cos^4theta=sin^2theta+bigg(cos^2thetabigg)^2=sin^2theta+bigg(1-sin^2thetabigg)^2=dots$$
answered Aug 30 '15 at 0:13
John JoyJohn Joy
6,29911727
answered Aug 30 '15 at 0:13
John JoyJohn Joy
6,29911727
answered Aug 30 '15 at 0:13
John JoyJohn Joy
6,29911727
answered Aug 30 '15 at 0:13
John JoyJohn Joy
6,29911727
6,29911727
$begingroup$
could you elaborate please?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:14
$begingroup$
Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
$endgroup$
– John Joy
Aug 30 '15 at 0:20
$begingroup$
$sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:26
$begingroup$
What do you get when you expand $(1-sin^2theta)^2$?
$endgroup$
– John Joy
Aug 30 '15 at 0:28
$begingroup$
$1- sin^4(theta) or cos^4(theta)$
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:36
|
show 9 more comments
$begingroup$
could you elaborate please?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:14
$begingroup$
Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
$endgroup$
– John Joy
Aug 30 '15 at 0:20
$begingroup$
$sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:26
$begingroup$
What do you get when you expand $(1-sin^2theta)^2$?
$endgroup$
– John Joy
Aug 30 '15 at 0:28
$begingroup$
$1- sin^4(theta) or cos^4(theta)$
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:36
$begingroup$
could you elaborate please?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:14
$begingroup$
could you elaborate please?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:14
$begingroup$
Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
$endgroup$
– John Joy
Aug 30 '15 at 0:20
$begingroup$
Yes, in order to prove that $sin^2(theta)+cos^4(theta)=cos^2(theta)+sin^4(theta)$ you need to put some thought into it.
$endgroup$
– John Joy
Aug 30 '15 at 0:20
$begingroup$
$sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:26
$begingroup$
$sin^2(theta) becomes sin^4(theta)$ after putting the expression to the power of two, but happens to the 1-?
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:26
$begingroup$
What do you get when you expand $(1-sin^2theta)^2$?
$endgroup$
– John Joy
Aug 30 '15 at 0:28
$begingroup$
What do you get when you expand $(1-sin^2theta)^2$?
$endgroup$
– John Joy
Aug 30 '15 at 0:28
$begingroup$
$1- sin^4(theta) or cos^4(theta)$
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:36
$begingroup$
$1- sin^4(theta) or cos^4(theta)$
$endgroup$
– Sunny Mann
Aug 30 '15 at 0:36
|
show 9 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1414134%2fprove-sin2-theta-cos4-theta-cos2-theta-sin4-theta%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Take the last step, and argue in reverse. e.g. $$sin^2theta-cos^2theta = sin^2theta-cos^2thetaimpliessin^2theta-cos^2theta=(sin^2theta+cos^2 theta)(sin^2theta-cos^2theta)=dots$$
$endgroup$
– John Joy
Aug 30 '15 at 0:23