Question : Simplify the following: $\sin2x+2\sin4x+\sin6x$
Option 1: $4\cos^2x\sin4x$
Option 2: $4\cos^2x\sin x$
Option 3: $2\cos^2x\sin4 x$
Option 4: $4\sin^2 x\sin4 x$
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Correct Answer: $4\cos^2x\sin4x$
Solution : Given: $\sin2x+2\sin4x+\sin6x$ $=\sin2x+\sin6x+2\sin4x$ Using the formula: $\sin C+\sin D=2\sin\frac{(C+D)}{2}\cos\frac{(C–D)}{2}$, we get: $= 2\sin\frac{(6x+2x)}{2}\cos\frac{(6x–2x)}{2}+2\sin4x$ $= 2\sin4x\cos2x+2\sin4x$ $=2\sin4x(\cos2x+1)$ Using formula: $2\cos^2x=\cos2x+1$, we get: $=2\sin4x(2\cos^2x)$ $= 4\cos^2x\sin4x$ Hence, the correct answer is $4\cos^2x\sin4x$.
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Question : Simplify the following: $\frac{\cos x-\sqrt{3} \sin x}{2}$
Option 1: $\cos \left(\frac{\pi}{3}-x\right)$
Option 2: $\sin \left(\frac{\pi}{3}+x\right)$
Option 3: $\cos \left(\frac{\pi}{3}+x\right)$
Option 4: $\sin \left(\frac{\pi}{3}-x\right)$
Question : Simplify the following. $\frac{\sin^3 \alpha+\cos^3 \alpha}{\sin \alpha+\cos \alpha}$
Option 1: $1+\sin \alpha \cos \alpha$
Option 2: $\tan \alpha$
Option 3: $1-\sin \alpha \cos \alpha$
Option 4: $\sec \alpha$
Question : Simplify the following expression. $\frac{\sin \theta - 2 \sin ^3 \theta}{2 \cos ^3 \theta - \cos \theta}$
Option 1: $\tan \theta$
Option 2: $\sin \theta$
Option 3: $\sec \theta$
Option 4: $\cos \theta$
Question : If $\cos x+\sin x=\sqrt{2} \cos x$, what is the value of $(\cos x-\sin x)^2+(\cos x+\sin x)^2$?
Option 1: $2$
Option 2: $1$
Option 3: $0$
Option 4: $\frac{1}{\sqrt{2}}$
Question : Simplify the given expression. $\frac{1+\sin^4 \theta+\cos^4 \theta}{\cos^2 \theta+\sin^4 \theta}$
Option 1: 3
Option 2: 1
Option 3: 2
Option 4: 4
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