Proving $\sin^2x \cos^2y - \cos^2x \sin^2y \;\equiv\; \cos^2y - \cos^2x$

Question

Knowing

$$\sin^2\theta +\cos^2\theta \equiv 1$$

how would I prove: $$\sin^2x \cos^2y - \cos^2x \sin^2y \;\equiv\; \cos^2y - \cos^2x$$

Can I substitute the first equation to prove the second one? If so, how can I?

Please help.

Please read this tutorial on how to typeset mathematics on this site. — N. F. Taussig, Jan 15 '19 at 13:47
The right-hand side of the target relation has only cosines in it. Can you use the Pythagorean identity to get rid of the sines on the left-hand side? — Blue, Jan 15 '19 at 13:47
See also https://math.stackexchange.com/questions/175143/prove-sinab-sina-b-sin2a-sin2b — lab bhattacharjee, Jan 15 '19 at 13:54

Cuhrazatee · Accepted Answer · 2019-02-18T17:44:23.730

1

Notice that the right hand side only has cosines in it. Try replacing all of the $\sin^2(x)$ and $\sin^2(y)$ with $1-\cos^2(x)$ and $1-\cos^2(y)$ respectively. If you simplify, you will see the right hand side.

edited Feb 18 '19 at 17:44

answered Jan 15 '19 at 14:40

Cuhrazatee

524

score 0 · Answer 2 · answered Jan 15 '19 at 14:54

Yes you can substitute the first equation into the second one.

$\sin^2\theta +\cos^2\theta \equiv 1 \implies \sin^2\theta = 1-\cos^2\theta$

Substitute to get,

$\begin{eqnarray} \sin^2x \cos^2y - \cos^2x \sin^2y &=& ((1-\cos^2(x)) (\cos^2y)) - ((\cos^2x) (1-\cos^2(y)) \\\\ &=& \cos^2y - ((\cos^2y)(\cos^2x)) - \cos^2x + ((\cos^2x)(\cos^2y)) \\\\ &=& \cos^2y - \cos^2x \end{eqnarray}$

score 0 · Answer 3 · answered Jan 15 '19 at 15:31

0

$\sin^2x \cos^2y - \cos^2x \sin^2y \\ = \sin^2x \cos^2y + (\cos^2x \cos^2y - \cos^2x \cos^2y) - \cos^2x \sin^2y \\ = (\sin^2x + \cos^2x) \cos^2y - \cos^2x (\cos^2y + \sin^2y) \\ = \cos^2y - \cos^2x$

answered Jan 15 '19 at 15:31

gandalf61

15,326

Proving $\sin^2x \cos^2y - \cos^2x \sin^2y \;\equiv\; \cos^2y - \cos^2x$

3 Answers3