Trigonometric Functions — MATHS STD 11 Science — Question

2. $\frac32$

Solution:

$\frac{2\pi}{3}=120^\circ$

Let $\text{f(x)}=\sin^2(90+30+\text{x})+\sin^2(90+30-\text{x})$

$=[\cos(30+\text{x})]^2+[\cos(30=\text{x})]^2$ $[\text{Using }\sin(90+\text{A})=\cos\text{A}]$

$=\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2+\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2$

$=\frac{\sqrt{3}}{2}\cos^2\text{x}-\frac14\sin^2\text{x}-\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}+\frac34\cos^2\text{x}+\frac14\sin^2\text{x}+\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}$

$=\frac{3}{2}\cos^2\text{x}-\frac12\sin^2\text{x}$

$=\frac{3}{2}(1-\sin^2\text{x})+\frac12\sin^2\text{x}$

$=\frac32-\frac{3}{2}\sin^2\text{x}+\frac12\sin^2\text{x}$

$=\frac32-\sin^2\text{x}$.

For f(x) to be maximum, $\sin^2\text{x}$ must have minimum value, which is 0.

$\therefore\frac32$ is the maximum value of f(x).