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Here is the equation:

$$\sin^2(a+b)+\sin^2(a-b)=1-\cos(2a)\cos(2b)$$

Following from comment help,

$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$

$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \cos^2 b + \cos^2 a \sin^2 b$$

I am stuck here, how do I proceed from here?

Edit: from answers I understand how to prove,but how to prove from where I am stuck?

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    $\begingroup$ Use the formulas for $\sin(x\pm y)$ and $\cos(2x)$ (look them up) and check that the two sides are equal. $\endgroup$
    – user228113
    Mar 22, 2017 at 10:41
  • $\begingroup$ @G.Sassatelli trying $\endgroup$
    – Fawad
    Mar 22, 2017 at 10:44
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    $\begingroup$ Post your attempt. And also, don't post images of equations, use mathjax. Really, as a long time user, you shoud know better than to post a low quality question like this. $\endgroup$
    – 5xum
    Mar 22, 2017 at 10:46
  • $\begingroup$ The last passage is wrong: $\sin^2a+\sin^2b\ne1$ $\endgroup$
    – user228113
    Mar 22, 2017 at 10:57
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    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Mar 22, 2017 at 11:13

8 Answers 8

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Well, let's the start the manipulation at the left hand side. Using the identity $$\sin^2\theta=\frac{1-\cos 2\theta}{2}$$ we get $$ \begin{align} \sin^2(a+b)+\sin^2(a-b)&=\frac{1-\cos(2a+2b)}{2}+\frac{1-\cos(2a-2b)}{2}\\ &=1-\frac{1}{2}\bigg[\cos(2a+2b)+\cos(2a-2b)\bigg]\\ &=1-\frac{1}{2}\bigg[(\cos 2a\cos 2b-\sin 2a\sin 2b)\\ &\qquad\qquad\qquad +(\cos 2a\cos 2b+\sin 2a\sin 2b)\bigg]\\ &=1-\frac{1}{2}\bigg[2\cos 2a\cos 2b\bigg]\\ &=1-\cos 2a\cos 2b. \end{align} $$

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  • $\begingroup$ i.stack.imgur.com/CEU3Q.jpg $\endgroup$
    – Fawad
    Mar 22, 2017 at 11:57
  • $\begingroup$ @Fawad What is the problem? why you paste that image? $\endgroup$
    – Juniven
    Mar 22, 2017 at 12:34
  • $\begingroup$ Your answer has mathjax problem don't know if it could be improved,it would be better. $\endgroup$
    – Fawad
    Mar 22, 2017 at 12:36
  • $\begingroup$ @Fawad Ah okay, nice pointing out the error. I will fix it and please take a review if it is already okay. $\endgroup$
    – Juniven
    Mar 22, 2017 at 12:38
  • $\begingroup$ @Fawad Done editing. Is it already okay? $\endgroup$
    – Juniven
    Mar 22, 2017 at 12:41
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On the LHS, you have $$2s_a^2c_b^2+2c_a^2s_b^2$$ after grouping.

On the RHS,

$$1-(c_a^2-s_a^2)(c_b^2-s_b^2)=(c_a^2+s_a^2)(c_b^2+s_b^2)-(c_a^2-s_a^2)(c_b^2-s_b^2)=2s_a^2c_b^2+2c_a^2s_b^2.$$

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  • $\begingroup$ Got,thanks alot $\endgroup$
    – Fawad
    Mar 22, 2017 at 11:06
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HINT:

Use Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ on

$$\sin^2(A+B)+\sin^2(A-B)=1-\{\cos^2(A+B)-\sin^2(A-B)\}$$

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On RHS we have, $$ 1-\cos(2a)\cos(2b) = 1-\cos^2a\cos^2b+\cos^2a\sin^2b+\sin^2a\cos^2b-\sin^2a\sin^2b = (\cos^2a\sin^2b+\sin^2a\cos^2b)+1-\cos^2a+\sin^b\cos^2a-\sin^2a\sin^2b = (\cos^2a\sin^2b+\sin^2a\cos^2b)+\sin^2b\cos^2a+\sin^2a-\sin^2a\sin^2b = 2(\cos^2a\sin^2b+\sin^2a\cos^2b) = LHS $$ Hence proved

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    $\begingroup$ Try using, e.g., $\cos x$ for $\cos x$ instead of $cos x$ for $cos x$. $\endgroup$
    – Shaun
    Mar 22, 2017 at 11:40
  • $\begingroup$ @Shaun edit done $\endgroup$
    – sgrmshrsm7
    Mar 22, 2017 at 11:48
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    $\begingroup$ @Shaun, thanks for guiding me. Thanks a lot $\endgroup$
    – sgrmshrsm7
    Mar 22, 2017 at 11:49
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We have $$ \begin{align} \sin^2(a+b)+\sin^2(a-b)&=\frac{1-\cos(2(a+b))}{2}+\frac{1-\cos(2(a-b))}{2}\\ &=1-\frac{\color{red}{\cos(2a+2b)}+\color{blue}{\cos(2a-2b)}}{2}\\ &=1-\frac{\color{red}{\cos 2a\cos 2b-\sin 2a\sin 2b}+ \color{blue}{\cos 2a\cos 2b+\sin 2a\sin 2b}}{2}\\ &=1-\frac{2\cos 2a\cos 2b}{2}\\ &=1-\cos 2a\cos 2b. \end{align} $$

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Other approach: let $u=a+b,v=a-b$.

Then

$$\begin{align}\cos(u+v)\cos(u-v)&=(\cos u\cos v-\sin u\sin v)\cdot(\cos u\cos v+\sin u\sin v)\\ &=\cos^2u\cos^2v-\sin^2u\sin^2v\\ &=(1-\sin^2u)(1-\sin^2v)-\sin^2u\sin^2v\\ &=1-\sin^2u-\sin^2v.\end{align}$$

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Given that $$ \cos(u) = \frac{e^{iu} + e^{-iu}}{2} , \sin(u) = \frac{e^{iu} - e^{-iu}}{2i} \text{ , and } e^{2u}+ e^{-2u} = \left(e^{u} - e^{-u}\right)^2 + 2,$$

we prove the identity by

$$\begin{align} 1 - \cos(2a) \cos(2b) & = 1 - \frac{ \left( e^{2ia} + e^{-2ia} \right) \left(e^{2ib} + e^{-2ib} \right)}{4}\\ & = \frac{e^{2i(a + b)} + e^{-2i(a + b)} + e^{2i(a - b)} + e^{-2i(a - b)} - 4}{-4}\\ & = \frac{ \left( e^{i(a + b)} - e^{-i(a + b)} \right)^2 + 2 + \left( e^{i(a - b)} - e^{-i(a - b)} \right)^2 + 2 - 4}{4i^2} \\ & = \sin^2(a+b) + \sin^2(a - b)\end{align} $$

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$\sin^2(a+b)+\sin^2(a−b) $

$=\sin^2a\cos^2b+\cos^2a\sin^2b+\sin^2a\cos^2b+\cos^2a\sin^2b $

$=2(\sin^2a\cos^2b+\cos^2a\sin^2b) $

$=2\left ({\left (\dfrac{(1-\cos(2a)}{2}\right )(\cos^2b)+\left (\dfrac{\cos(2a)+1}{2}\right )(\sin^2b)}\right ) $

$=1+\cos(2a)\sin^2b-\cos(2a)\cos^2b $

$=1-\cos(2a)\cos(2b) $

Hence proved.

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