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Question Number 27296 by julli deswal last updated on 04/Jan/18

$$\frac{{\sin }^{2}3A}{{\sin }^{2}A}-\frac{{\cos }^{2}3A}{{\cos }^{2}A}=$$

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

$$=\frac{{\left(sin\left(3a\right)cosa\right)}^2-{\left(cos\left(3a\right)sina\right)}^2}{{\left(sinacosa\right)}^2}$$$$=\frac{(sin\left(3a\right)cosa-cos\left(3a\right)sina)(sin\left(3a\right)cosa+cos\left(3a\right)sina)}{{\left(sinacosa\right)}^2}$$$$=\frac{sin\left(2a\right).sin\left(4a\right)}{{\left(\frac{1}{2}sin\left(2a\right)\right)}^2}$$$$=4\frac{sin\left(4a\right)}{sin\left(2a\right)}=\frac{8sin\left(2a\right)cos\left(2a\right)}{sin\left(2a\right)}=8cos\left(2a\right).$$

Answered by Giannibo last updated on 04/Jan/18

$$$$$$$$$$\frac{{\sin }^2(2\mathrm{A}+\mathrm{A})}{{\sin }^2\mathrm{A}}-\frac{\cos {}^2(2\mathrm{A}+\mathrm{A})}{\cos {}^2\mathrm{A}}=$$$$\frac{{(\sin 2\mathrm{A}\cdot \cos \mathrm{A}+\cos 2\mathrm{A}\cdot \sin \mathrm{A})}^2}{\sin {}^2\mathrm{A}}-\frac{{(\cos 2\mathrm{A}\cdot \cos \mathrm{A}-\sin 2\mathrm{A}\cdot \sin \mathrm{A})}^2}{\cos {}^2\mathrm{A}}$$$$\frac{{(2\sin \mathrm{A}\cdot {\cos }^2\mathrm{A}+(\cos {}^2\mathrm{A}-\sin {}^2\mathrm{A})\sin \mathrm{A})}^2}{\sin {}^2\mathrm{A}}-\frac{{((\cos {}^2\mathrm{A}-\sin {}^2\mathrm{A})\cos \mathrm{A}-2\sin \mathrm{A}\cdot \cos \mathrm{A}\cdot \sin \mathrm{A})}^2}{\cos {}^2\mathrm{A}}=$$$$\frac{\sin {}^2\mathrm{A}{({2\cos }^2\mathrm{A}+\cos {}^2\mathrm{A}-\sin {}^2\mathrm{A})}^2}{\sin {}^2\mathrm{A}}-\frac{\cos {}^2\mathrm{A}\cdot {(\cos {}^2\mathrm{A}-\sin {}^2\mathrm{A}-2\sin {}^2\mathrm{A})}^2}{\cos {}^2\mathrm{A}}=$$$${(3\cos {}^2\mathrm{A}-\sin {}^2\mathrm{A})}^2-{(\cos {}^2\mathrm{A}-3\sin {}^2\mathrm{A})}^2=$$$$9\cos {}^4\mathrm{A}-6\cos {}^2\mathrm{Asin}{}^2\mathrm{A}+\sin {}^4\mathrm{A}-\cos {}^4\mathrm{A}+6\cos {}^2\mathrm{Asin}{}^2\mathrm{A}-9\sin {}^4\mathrm{A}=$$$$8\cos {}^4\mathrm{A}-8\sin {}^4\mathrm{A}=$$$$8(\cos {}^2\mathrm{A}-\sin {}^2\mathrm{A})(\cos {}^2\mathrm{A}+\sin {}^2\mathrm{A})=$$$$8\cos 2\mathrm{A}$$

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