How do you prove that $1 - cos(5theta)cos(3theta) - sin(5theta)sin(3theta) = 2sin^2theta$ ?

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Answer 1 · 2017-03-21T23:44:04

Factor.

$1 - (cos5thetacos3theta + sin5thetasin3theta) = 2sin^2theta$

Note that $cosAcosB + sinAsinB = cos(A - B)$ .

$1 - (cos(5theta - 3theta)) = 2sin^2theta$

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

Now use $cos(A + B) = cosAcosB - sinAsinB$ (because $cos2theta = cos(theta + theta)$ ) to find an expansion for $cos(2theta)$ .

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

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

Now apply $sin^2x + cos^2x = 1$ . This implies that $sin^2x = 1 - cos^2x$ .

$sin^2theta + sin^2theta = 2sin^2theta$

$2sin^2theta = 2sin^2theta$

$LHS = RHS$

Since both sides are equal for all values of $theta$ , this identity has been proved.

Hopefully this helps!

How do you prove that 1 - cos(5theta)cos(3theta) - sin(5theta)sin(3theta) = 2sin^2theta1 - cos(5theta)cos(3theta) - sin(5theta)sin(3theta) = 2sin^2theta?