Question
If $\theta<90^{\circ}$, find the value of: $\sin ^2 \theta+\cos ^2 \theta$
✓
Answer
Since $\theta<90^{\circ}$,
Consider $\theta=45^{\circ}$
$\therefore \sin ^2+\cos ^2$
$=\sin ^2 45^{\circ}+\cos ^2 45^{\circ}$
$=\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2$
$=\frac{1}{2}+\frac{1}{2}$
$=1$
Consider $\theta=45^{\circ}$
$\therefore \sin ^2+\cos ^2$
$=\sin ^2 45^{\circ}+\cos ^2 45^{\circ}$
$=\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2$
$=\frac{1}{2}+\frac{1}{2}$
$=1$
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