## How do you solve \(\sin ^{2} \theta-\cos ^{2} \theta=0 ?\)

Answer 1:

\(\frac{\pi}{4}+k \pi\)

\(\frac{3 \pi}{4}+k \pi\)

Explanation:

From the trig identity:

\(\cos ^{2} t-\sin ^{2} t=\cos 2 t\) we get:

\(\sin ^{2} t-\cos ^{2} t=-\cos 2 t=0\)

Unit circle gives 2 solutions:

\(\cos 2 t=0\) − →2t = pi/2 + 2kpi, \(and\) 2t = (3pi)/2 + 2kpi a. 2t = pi/ + 2kpi− →t = pi/4 + kpib. 2t = (3pi)/2 + 2kpi− →t = (3pi)/4 + kpi#

Answer 2:

Here’s an alternate answer. Recall the identity \(\sin ^{2} \theta+\cos ^{2} \theta=1\) If you rearrange for \(\cos ^{2} \theta,\) you should get \(\cos ^{2} \theta=1-\sin ^{2} \theta\).

Substituting, we have:

\(\sin ^{2} \theta-\left(1-\sin ^{2} \theta\right)=0\)

\(\sin ^{2} \theta-1+\sin ^{2} \theta=0\)

\(2 \sin ^{2} \theta=1\)

\(\sin ^{2} \theta=\frac{1}{2}\)

\(\sin \theta=\pm \frac{1}{\sqrt{2}}\)

Now consider the \(1-1-\sqrt{2}\) right triangle. This means that

\(\theta=\frac{\pi}{4}+2 \pi n, \frac{3 \pi}{4}+2 \pi n, \frac{5 \pi}{4}+2 \pi n \text { and } \frac{7 \pi}{4}+2 \pi n\)Note the period of the sine function is \(2π\).

Hopefully this helps!