If left(frac{sin 3 theta}{sin theta}right)^2- | vedclass

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Question

(A) We have the expression $\left(\frac{\sin 3 	heta}{\sin 	heta}ight)^2-\left(\frac{\cos 3 	heta}{\cos 	heta}ight)^2$. Using $\sin 3 	heta =

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$4: 1$

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(A) We have the expression $\left(\frac{\sin 3 	heta}{\sin 	heta}ight)^2-\left(\frac{\cos 3 	heta}{\cos 	heta}ight)^2$. Using $\sin 3 	heta = 3 \sin 	heta - 4 \sin^3 	heta$ and $\cos 3 	heta = 4 \cos^3 	heta - 3 \cos 	heta$,we get: $\left(\frac{3 \sin 	heta - 4 \sin^3 	heta}{\sin 	heta}ight)^2 - \left(\frac{4 \cos^3 	heta - 3 \cos 	heta}{\cos 	heta}ight)^2$ $= (3 - 4 \sin^2 	heta)^2 - (4 \cos^2 	heta - 3)^2$ $= (3 - 4(1 - \cos^2 	heta))^2 - (4 \cos^2 	heta - 3)^2$ $= (4 \cos^2 	heta - 1)^2 - (4 \cos^2 	heta - 3)^2$ Using $x^2 - y^2 = (x - y)(x + y)$: $= ((4 \cos^2 	heta - 1) - (4 \cos^2 	heta - 3))((4 \cos^2 	heta - 1) + (4 \cos^2 	heta - 3))$ $= (2)(8 \cos^2 	heta - 4) = 16 \cos^2 	heta - 8$ $= 8(2 \cos^2 	heta - 1) = 8 \cos 2 	heta$. Comparing this with $a \cos b 	heta$,we get $a = 8$ and $b = 2$. Therefore,$a: b = 8: 2 = 4: 1$.

If $\left(\frac{\sin 3 \theta}{\sin \theta}\right)^2-\left(\frac{\cos 3 \theta}{\cos \theta}\right)^2=a \cos b \theta$,then $a: b=$

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