Question

Maximum and Minimum value of ${\sin }^2(120+\theta )+{\sin }^2(120-\theta )$ are

• A. $\frac{3}{2},\frac{1}{2}$
• B. $\frac{1}{2},0$
• C. $\frac{3}{2},0$
• D. $\frac{3}{2},\frac{1}{3}$

Answer 1

Answer: (A)

$\frac{3}{2},\frac{1}{2}$

$=1+{\sin }^2(120+\theta )-{\cos }^2(120-\theta )$

$=1+\left(\sin \right(120+\theta )+\cos (120-\theta \left)\right)\left(\sin \right(120+\theta )-\cos (120-\theta \left)\right)$

$=1+(\sin 120\cos \theta +\cos 120\sin \theta +\cos 120\cos \theta +\sin 120\sin \theta )\times (\sin 120\cos \theta +\cos 120\sin \theta -\cos 120\cos \theta -\sin 120\sin \theta )$

$=1+(\frac{\sqrt{3}\cos \theta }{2}-\frac{\sin \theta }{2}-\frac{\cos \theta }{2}+\frac{\sqrt{3}\sin \theta }{2})(\frac{\sqrt{3}\cos \theta }{2}-\frac{\sin \theta }{2}+\frac{\cos \theta }{2}-\frac{\sqrt{3}\sin \theta }{2})$

$=1+\left(\frac{\sqrt{3}-1}{2}\right(\cos \theta +\sin \theta \left)\right)\left(\frac{\sqrt{3}+1}{2}\right(\cos \theta -\sin \theta \left)\right)$

$=1+\frac{3-1}{4}({\cos }^2\theta -{\sin }^2\theta )$
$=1+\frac{\cos 2\theta }{2}$
Value is max when $\cos 2\theta =1$
$=\frac{3}{2}$
Value is min when $\cos 2\theta =-1$
$=\frac{1}{2}$