frac{sin 5 theta}{sin theta} is equal to | vedclass

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Question

(A) We know that $\sin 5 	heta = 5 \sin 	heta - 20 \sin ^3 	heta + 16 \sin ^5 	heta$. Dividing by $\sin 	heta$ (assuming $\sin 	heta 
eq 0$): $

Vedclass · Accepted Answer

$16 \cos ^4 	heta-12 \cos ^2 	heta+1$

Vedclass · Answer

(A) We know that $\sin 5 	heta = 5 \sin 	heta - 20 \sin ^3 	heta + 16 \sin ^5 	heta$. Dividing by $\sin 	heta$ (assuming $\sin 	heta 
eq 0$): $\frac{\sin 5 	heta}{\sin 	heta} = 5 - 20 \sin ^2 	heta + 16 \sin ^4 	heta$. Using the identity $\sin ^2 	heta = 1 - \cos ^2 	heta$: $= 5 - 20(1 - \cos ^2 	heta) + 16(1 - \cos ^2 	heta)^2$ $= 5 - 20 + 20 \cos ^2 	heta + 16(1 - 2 \cos ^2 	heta + \cos ^4 	heta)$ $= -15 + 20 \cos ^2 	heta + 16 - 32 \cos ^2 	heta + 16 \cos ^4 	heta$ $= 16 \cos ^4 	heta - 12 \cos ^2 	heta + 1$.

$A$. $Pa \cdot s$	$(i)$ $[L^2 \ T^{-2} \ K^{-1}]$
$B$. $N \cdot m \cdot K^{-1}$	$(ii)$ $[M \ L \ T^{-3} \ K^{-1}]$
$C$. $J \cdot kg^{-1} \cdot K^{-1}$	$(iii)$ $[M \ L^{-1} \ T^{-1}]$
$D$. $W \cdot m^{-1} \cdot K^{-1}$	$(iv)$ $[M \ L^2 \ T^{-2} \ K^{-1}]$

$\frac{\sin 5 \theta}{\sin \theta}$ is equal to

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