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Question Number 29111 by gyugfeet last updated on 04/Feb/18

$${\cos }^{3}x.\sin {}^{2}x=1/16(2\cos x-\cos 3x-\cos 5x)$$

Commented by abdo imad last updated on 04/Feb/18

$$anothermethod$$$${cos}^{3}x{sin}^{2}x=cosx{\left(cosxsinx\right)}^2=\frac{1}{4}cosx{\left(sin\left(2x\right)\right)}^2$$$$=\frac{1}{4}cosx\left(\frac{1-cos\left(4x\right)}{2}\right)=\frac{1}{8}(cosx-cosxcos\left(4x\right))but$$$$cosxcos\left(4x\right)=\frac{1}{2}(cos\left(5x\right)+cos\left(3x\right))\Rightarrow$$$${cos}^{3}x{sin}^{2}x=\frac{1}{8}(cosx-\frac{1}{2}cos\left(3x\right)-\frac{1}{2}cos\left(5x\right))$$$$=\frac{1}{16}(2cosx-cos\left(3x\right)-cos\left(5x\right)).$$

Answered by ajfour last updated on 04/Feb/18

$$\cos 3x=4\cos {}^{3}x-3\cos x$$$$2\sin {}^{2}x=1-\cos 2x$$$$so$$$$\left(4\cos {}^{3}x\right)\left(2\sin {}^{2}x\right)=(3\cos x+\cos 3x)(1-\cos 2x)$$$$=3\cos x-3\cos x\cos 2x$$$$+\cos 3x-\cos 3x\cos 2x$$$$=3\cos x-\frac{3}{2}(\cos x+\cos 3x)$$$$+\cos 3x-\frac{1}{2}(\cos x+\cos 5x)$$$$=\cos x-\frac{1}{2}\cos 3x-\frac{1}{2}\cos 5x$$$$\Rightarrow \cos {}^{3}x\sin {}^{2}x=$$$$\frac{1}{16}(2\cos x-\cos 3x-\cos 5x).$$

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