Question

I: lf $y=\sin x\sin 2x\sin 3x$ then $y_n=\frac{1}{4}\left[4^n\sin \right(\frac{n\pi }{2}+4x)-6^n\sin (\frac{n\pi }{2}+6x)+2^n\sin (\frac{n\pi }{2}+2x\left)\right]$
II: lf $y=\sin 2x\sin 3x\sin 4x$ then $y_n=\frac{1}{4}[-9^n\sin (\frac{n\pi }{2}+9x)+5^n\sin (\frac{n\pi }{2}+5x)$ $+3^n\sin (\frac{n\pi }{2}+3x)+\sin (\frac{n\pi }{2}+x)]$
Which of the following is correct?

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II true

Answer 1

Answer: (C)

Both I and II are true

$I.y=\frac{1}{2}\left(2\sin x\sin 2x\right)\sin 3x=\frac{1}{2}(\cos x-\cos 3x)\sin 3x$

$y=\frac{1}{4}(\sin 4x+\sin 2x-\sin 6x)$

$y^{{}^{\prime }}=\frac{1}{4}(4\sin (\frac{\pi }{2}+4x)+2\sin (\frac{\pi }{2}+2x)-6\sin (\frac{\pi }{2}+6x))$

$y^{{}^{\prime \prime }}\frac{1}{4}\left(4^2\sin \right(\pi +4x)+2^2\sin (\pi +2x)-6^2\sin (\pi +6x\left)\right)$

$y^n=\frac{1}{4}\left(4^n\sin \right(\frac{n\pi }{2}+4x)+2^n\sin (\frac{n\pi }{2}+2x)-6^n\sin (\frac{n\pi }{2}+6x\left)\right)$

$II.y=\frac{1}{2}\left(2\sin 2x\sin 3x\sin 4x\right)=\frac{1}{2}(\cos x-\cos 5x)\sin 4x$

$y=\frac{1}{4}\left(2\cos x\sin 4x\right)=\frac{1}{4}(\sin 5x+\sin 3x-\sin 9x+\sin x)$

$y^n=\frac{1}{4}(5^n\sin (\frac{n\pi }{2}+5x)+3^n\sin (\frac{n\pi }{2}+3x)-9^n\sin (\frac{n\pi }{2}+9x)+\sin (\frac{n\pi }{2}+x))$

Hence, both statements are correct.