frac{{sin 3theta - cos 3theta }}{{sin theta + | vedclass

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Question

(A) Let the given expression be $E = \frac{{\sin 3	heta - \cos 3	heta }}{{\sin 	heta + \cos 	heta }} + 1$.Using the identities $\sin 3	heta = 3\s

Vedclass · Accepted Answer

$2\sin 2	heta $

Vedclass · Answer

(A) Let the given expression be $E = \frac{{\sin 3	heta - \cos 3	heta }}{{\sin 	heta + \cos 	heta }} + 1$.Using the identities $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$ and $\cos 3	heta = 4\cos^3 	heta - 3\cos 	heta$,we have:$N = \sin 3	heta - \cos 3	heta = (3\sin 	heta - 4\sin^3 	heta) - (4\cos^3 	heta - 3\cos 	heta)$$N = 3(\sin 	heta + \cos 	heta) - 4(\sin^3 	heta + \cos^3 	heta)$Using the sum of cubes formula $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$:$N = 3(\sin 	heta + \cos 	heta) - 4(\sin 	heta + \cos 	heta)(\sin^2 	heta - \sin 	heta \cos 	heta + \cos^2 	heta)$Since $\sin^2 	heta + \cos^2 	heta = 1$,we get:$N = (\sin 	heta + \cos 	heta) [3 - 4(1 - \sin 	heta \cos 	heta)]$Now,divide by $D = \sin 	heta + \cos 	heta$:$\frac{N}{D} = 3 - 4 + 4\sin 	heta \cos 	heta = 4\sin 	heta \cos 	heta - 1$Adding $1$ to the expression:$E = (4\sin 	heta \cos 	heta - 1) + 1 = 4\sin 	heta \cos 	heta$Using the double angle identity $2\sin 	heta \cos 	heta = \sin 2	heta$:$E = 2(2\sin 	heta \cos 	heta) = 2\sin 2	heta$.

$\frac{{\sin 3\theta - \cos 3\theta }}{{\sin \theta + \cos \theta }} + 1 = $

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