Question

Use the sum and difference formulas or a double angle formula to rewrite theexpressions as sin, cos or tan(a) cos^2x - ½(b) 6 sin x cos x

Answer 1

We have the following trigonometric identities:

$\cos2x=2\cos\placeholder{⬚}^2x-1$$\sin2x=2\sin x\cos x$

a)

From the first identity above we have that:

$\cos\placeholder{⬚}^2x=\frac{\cos2x}{2}+\frac{1}{2}$

Plugging this in the expression given we have:

$\begin{gathered} \cos\placeholder{⬚}^2x-\frac{1}{2}=\frac{\cos2x}{2}+\frac{1}{2}-\frac{1}{2} \\ \cos\placeholder{⬚}^2x-\frac{1}{2}=\frac{1}{2}\cos2x \end{gathered}$

Therefore:

$\cos\placeholder{⬚}^2x-\frac{1}{2}=\frac{1}{2}\cos2x$

b)

From the second identity given at the beginning we have:

$\sin x\cos x=\frac{1}{2}\sin2x$

Plugging this in the second expression we have:

$\begin{gathered} 6\sin x\cos x=6(\frac{1}{2}\sin2x) \\ 6\sin x\cos x=3\sin2x \end{gathered}$

Therefore:

$6\sin x\cos x=3\sin2x$