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!