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Answer 1 · 2017-11-20T02:51:12

$cos^2(C)-cos^2(D) = sin^2(D) - sin^2(C)$

A common formula used all the time in trigonometry is

$sin^2(theta)+cos^2(theta)=1$

Subtracting the $sin^2(theta)$ from both sides gives

$cos^2(theta)=1-sin^2(theta)$

This means that the formula $cos^2(C)-cos^2(D)$ can be expressed as

$cos^2(C)-cos^2(D) = (1-sin^2(C))-(1-sin^2(D))$

$=-sin^2(C)+sin^2(D)$

$=sin^2(D) - sin^2(C)$

Answer 2 · 2017-11-20T02:51:16

$-sin^2C+sin^2D$

If you meant to ask how to turn the expression into something in terms of a sine functions this is how:

We know:

$sin^2x+cos^2x=1$

If we subtract $sin^2x$ from both sides we get:

$sin^2x+cos^x-sin^2x=1-sin^2x$

$cos^2x=1-sin^2x$

Therefore:

$1-sin^2C-(1-sin^2D)=1-sin^2C-1+sin^2D=cancel1-sin^2Ccancel-1+sin^2D=-sin^2C+sin^2D$