Question

The minimum and maximum values of $sin(x+\frac{\pi }{4})sin(x-\frac{\pi }{4})$ are

• A. $-1,1$
• B. $-\frac{1}{2},\frac{1}{2}$
• C. $-\frac{3}{2},\frac{3}{2}$
• D. $-\frac{1}{4},\frac{1}{4}$

Answer 1

The correct option is B $-\frac{1}{2},\frac{1}{2}$

Given

$\sin (x+\frac{\pi }{4})\cdot \sin (x-\frac{\pi }{4})$

we know that

$\sin (A+B)\cdot \sin (A-B)={\sin }^{2}A-{\sin }^{2}B$

$\Rightarrow \sin (x+\frac{\pi }{4})\cdot \sin (x-\frac{\pi }{4})={\sin }^{2}x-{\sin }^2\frac{\pi }{4}$

$={\sin }^{2}x-{\left(\frac{1}{\sqrt{2}}\right)}^2$

$={\sin }^{2}x-\frac{1}{2}$

But

$\sin x$ lies in $[-1,1]$

$\Rightarrow -1\le \sin x\le +1$

So,

$0\le {\sin }^{2}x\le 1$

$\Rightarrow 0-\frac{1}{2}\le {\sin }^{2}x-\frac{1}{2}\le 1-\frac{1}{2}$

$\Rightarrow -\frac{1}{2}\le {\sin }^{2}x-\frac{1}{2}\le \frac{1}{2}$

The function $sin(x+\frac{\pi }{4})\cdot \sin (x-\frac{\pi }{4})$ has the

minimum values: $\frac{-1}{2}$

maximum values: $\frac{1}{2}$