Question

If $x\sin \theta +y\sin 2\theta +z\sin 3\theta =$ $\sin 4\theta ,(\theta \nleqq n\pi )$, then $8{\cos }^3\theta -4z{\cos }^2\theta -2(y+2)\cos \theta =$

• A. $x-y$
• B. $x-z$
• C. $y-z$
• D. $0$

Answer 1

The correct option is B $x-z$

$x\sin \theta +y\sin 2\theta +z\sin 3\theta =\sin 4\theta$

$sox\sin \theta +2y\sin \theta \cos \theta +z(3\sin \theta -4{\sin }^3\theta )=2\left(2\sin \theta \cos \theta \right)(2{\cos }^2\theta -1)$

now we cancel $\sin \theta$ and put ${\sin }^2\theta =1-{\cos }^2\theta$ and simplify

$\Rightarrow \sin \theta (x+2y\cos \theta +z(3-4{\sin }^2\theta \left)\right)=4\left(\sin \theta \right)\cos \theta (2{\cos }^2\theta -1)$

$\Rightarrow (x+2y\cos \theta +z(3-4{\sin }^2\theta \left)\right)=4\cos \theta (2{\cos }^2\theta -1)$

$\Rightarrow (x+2y\cos \theta +z(3-4(1-{\cos }^2\theta )\left)\right)=4\cos \theta (2{\cos }^2\theta -1)$

$\Rightarrow (x+2y\cos \theta +z(3-4+4{\cos }^2\theta \left)\right)=4\cos \theta (2{\cos }^2\theta -1)$

$\Rightarrow x+2y\cos \theta +z(-1+4{\cos }^2\theta )=8{\cos }^3\theta -4\cos \theta$

$\Rightarrow x-z+2y\cos \theta +4z{\cos }^2\theta =8{\cos }^3\theta -4\cos \theta$

$\Rightarrow x-z=8{\cos }^3\theta -4\cos \theta -(2y\cos \theta +4z{\cos }^2\theta )$

$\Rightarrow x-z=8{\cos }^3\theta -4\cos \theta -2y\cos \theta -4z{\cos }^2\theta$

$\Rightarrow x-z=8{\cos }^3\theta -(4+2y)\cos \theta -4z{\cos }^2\theta$

$\Rightarrow x-z=8{\cos }^3\theta -2(2+y)\cos \theta -4z{\cos }^2\theta$

$\Rightarrow x-z=8{\cos }^3\theta -4z{\cos }^2\theta -2(2+y)\cos \theta$

$\therefore$ $8{\cos }^3\theta -4z{\cos }^2\theta -2(2+y)\cos \theta =x-z$