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

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Question

(A) Let $E = \frac{\sin 3	heta - \cos 3	heta}{\sin 	heta + \cos 	heta} + 1$. Using the identities $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$ a

Vedclass · Accepted Answer

$2\sin 2	heta$

Vedclass · Answer

(A) Let $E = \frac{\sin 3	heta - \cos 3	heta}{\sin 	heta + \cos 	heta} + 1$. Using the identities $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$ and $\cos 3	heta = 4\cos^3 	heta - 3\cos 	heta$: Numerator $N = \sin 3	heta - \cos 3	heta = (3\sin 	heta - 4\sin^3 	heta) - (4\cos^3 	heta - 3\cos 	heta) = 3(\sin 	heta + \cos 	heta) - 4(\sin^3 	heta + \cos^3 	heta)$. 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 have: $N = (\sin 	heta + \cos 	heta) [3 - 4(1 - \sin 	heta \cos 	heta)]$. Thus,$\frac{N}{\sin 	heta + \cos 	heta} = 3 - 4 + 4\sin 	heta \cos 	heta = 4\sin 	heta \cos 	heta - 1$. 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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