Question

Assertion :Statement I: The range of $f\left(x\right)=\sin (\frac{\pi }{5}+x)-\sin (\frac{\pi }{5}-x)-\sin (\frac{2\pi }{5}+x)+\sin (\frac{2\pi }{5}-x)$ is $[-1,1]$ Reason: Statement II: $\cos \frac{\pi }{5}-\cos \frac{2\pi }{5}=\frac{1}{2}$

• A. Statement I is true,Statement II is also true; Statement II is the correct explanation of Statement I
• B. Statement I is true,Statement II is also true; Statement II is not the correct explanation of Statement I
• C. Statement I is true,Statement II is false
• D. Statement I is false,Statement II is true

Answer 1

The correct option is A Statement I is true,Statement II is also true; Statement II is the correct explanation of Statement I

Reason:

$\cos \frac{\pi }{5}-\cos \frac{2\pi }{5}=\cos {36}^{\circ }-\cos {72}^{\circ }=\cos {36}^{\circ }-\sin {18}^{\circ }$

$=\frac{\sqrt{5}+1}{4}-\left(\frac{\sqrt{5}-1}{4}\right)=\frac{1}{2}$

Assertion:

$f\left(x\right)=\sin (\frac{\pi }{5}+x)-\sin (\frac{\pi }{5}-x)-\sin (\frac{2\pi }{5}+x)+\sin (\frac{2\pi }{5}-x)$

By using,

$\sin (A+B)+\sin (A-B)=2\sin A\cos B$

$f\left(x\right)=2\cos \frac{\pi }{5}\sin x-2\cos \frac{2\pi }{5}\sin x$

$=2\sin x[\cos \frac{\pi }{5}-\cos \frac{2\pi }{5}]\dots \left(fromreason\right)$

$=2\sin x\left[\frac{1}{2}\right]=\sin x$

Hence range is $[-1,1]$