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So I started by using $\sin 3A$ and $\cos 3A$ identities and then I added the lone $1$ to the trigonometric term. (Done in the picture below)

But after this I don't have any clue on how to proceed.

$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-\left(4 \cos ^{3} \theta-\sin \theta\right)}{\sin \theta+\cos \theta} + 1$$ $$=\frac{3 \sin \theta-4\sin^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta}{\sin \theta+\cos \theta} + 1$$ $$=\frac{3 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta+\sin \theta+\cos \theta}{\sin \theta+\cos \theta}$$ $$=\frac{4 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+4 \cos \theta}{\sin \theta+\cos \theta}$$ $$=\frac{4 \sin \theta+\cos \theta-\sin ^{3} \theta-\cos ^{3} \theta} {\sin \theta+\cos \theta} $$

Original image

The 2nd
  • 856
Shogun BS
  • 51

3 Answers3

4

Recall the cube sum identity: $x^3+y^3 = (x+y)(x^2-xy+y^2)$

Continuing from where you left off:

\begin{align}&\quad4\frac {\sin\theta+\cos\theta-\sin^3\theta-\cos^3\theta}{\sin\theta+\cos\theta}\\ &=4-4\frac {\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta}\\ &=4-4(\sin^2\theta - \sin\theta\cos\theta+\cos^2\theta)\\ &=4-4(1-\frac12\sin2\theta)\\ &=2\sin2\theta \end{align}

player3236
  • 16,413
2

$$\frac{\sin3x-\cos3x}{\sin x+\cos x}+1=\frac{\sin3x+\sin x+\cos x-\cos3x}{\sin x+\cos x}=\frac{\sin(2x+x)+\sin (2x-x)+\cos (2x-x)-\cos(2x+x)}{\sin x+\cos x}=\frac{\sin2x\cos x+\sin x\cos2x+\sin2x\cos x-\sin x\cos 2x+\cos 2x\cos x+ \sin 2x\sin x-(\cos 2x\cos x- \sin 2x\sin x)}{\sin x+\cos x}=\frac{2 \sin2x\cos x +2 \sin 2x\sin x}{\sin x+\cos x}=\frac{2 \sin2x(\cos x +\sin x)}{\sin x+\cos x}=2\sin2x$$

Lion Heart
  • 7,073
0

$$\sin3A-\cos3A$$ $$=\sqrt2\sin\left(3A-\dfrac\pi4\right)$$

$$=\sqrt2\sin\left(3A+\dfrac{3\pi}4-\pi\right)$$

$$=-\sqrt2\sin3\left(A+\dfrac\pi4\right)$$

Similarly, $\sin A+\cos A=\sqrt2\sin(?)$

Finally for $\sin B\ne0,$ $$\dfrac{\sin3B}{\sin B}=\dfrac{\sin B(3-4\sin^2B)}{\sin B}=3-2(1-\cos2B)$$

lab bhattacharjee
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