I want to find out value of this expression $$\cos^2 48°-\sin^2 12°$$ Just hint the starting step.Is there any any formula regarding $\cos^2 A-\sin^2 B$?
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@Jerry but there is not any formula I read like ($\cos A-\sin B$) – iostream007 May 12 '13 at 09:17
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Do you have any reason to believe it gets any simpler? – Mike May 12 '13 at 09:22
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Yes I believe because it was asked in a exam in 1994 – iostream007 May 12 '13 at 09:23
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$$\cos^{2}(48)-\sin^{2}(12)=\sin^{2}(42)-\sin^{2}(12)=\sin(54)\sin(30)=\dfrac{1}{2(\sqrt{5}-1)}$$
Here we used the formula-$$\sin^{2}(A)-\sin^{2}(B)=\sin(A+B)\sin(A-B)$$ and $\sin(54)=\dfrac{1}{\sqrt{5}-1}$
Shaswata
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I've got a formula : $$\cos(A+B).\cos(A-B)=\cos^2A-\sin^2B$$ so from this formula this question is now easy $$\cos^248-\sin^212$$ $$\cos60.\cos36$$ $$\frac{\sqrt{5}+1}{8}$$
iostream007
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