Question

A. Show

Answer 1

a. Recall the double angle identities:

$\sin^2x=\dfrac{1-\cos2x}2$

$\cos^2x=\dfrac{1+\cos2x}2$

Then

$\sin^2x\cos^2x=\dfrac{(1-\cos2x)(1+\cos2x)}4=\dfrac{1-\cos^22x}4$

Applying the identity again, we have

$\sin^2x\cos^2x=\dfrac{1-\frac{1+\cos4x}2}4=\dfrac{2-(1+\cos4x)}8=\dfrac{1-\cos4x}8$

as required.

b. Using the result from part (a),

$\sin^2x\cos^2x=\dfrac{1-\cos4x}8=\dfrac{2-\sqrt2}{16}$

$\implies\cos4x=\dfrac1{\sqrt2}$

$\implies4x=\dfrac\pi4+2n\pi\text{ or }4x=-\dfrac\pi4+2n\pi$

(where $n$ is any integer)

$\implies\boxed{x=\pm\dfrac\pi{16}+\dfrac{n\pi}2}$