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Prove the given trigonometric identity: $$\cos^6 \beta - \sin^6 \beta=\frac {1}{16} (15 \cos 2\beta + \cos 6\beta)$$

My Approach: \begin{align*} \text{L.H.S.} &=\cos^6 \beta - \sin^6 \beta\\ &=(\cos^2 \beta)^3 - (\sin^2 \beta)^3\\ &=\cos^32\beta+3\cos^2\beta\cdot\sin^2\beta\cdot\cos2\beta \end{align*}

Please help me to continue further.

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    $\begingroup$ Use : $$a^3-b^3=(a-b)(a^2+b^2+ab)$$ $\endgroup$ – Aakash Kumar Aug 27 '16 at 9:49
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$$(\cos^2\beta)^3 -(\sin^2\beta)^3=(\cos^2\beta-\sin^2\beta)(\cos^4\beta +\sin^4\beta +\cos^2\beta\sin^2\beta)$$ $$=\cos2\beta(1-\sin^2\beta\cos^2\beta)$$ $$=\cos2\beta\left(1-\frac{\sin^2 2\beta}{4}\right)$$ $$=\cos2\beta\left(1+\frac{\cos 4\beta -1}{8}\right)$$ Now use $2\cos A.\cos B=\cos(A+B) +\cos(A-B)$ $$=\frac18\left(7\cos2\beta +\cos 2\beta\cos 4\beta\right)$$ $$=\frac18\left(7\cos 2\beta +\frac{\cos 6\beta + \cos 2\beta}{2}\right)$$ $$=\frac{1}{16}\left(15\cos 2\beta + \cos 6 \beta\right)$$

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Using Euler Identity $$e^{ix}=\cos x+i\sin x,$$

$2\cos x=e^{ix}+e^{-ix},2i\sin x=e^{ix}-e^{-ix}$

$$64(\cos^6\beta-\sin^6\beta)=(2\cos\beta)^6+(2i\sin\beta)^6$$ $$=(e^{i\beta}+e^{-i\beta})^6+(e^{i\beta}-e^{-i\beta})^6$$

$$=2\left\{e^{i6\beta}+e^{-i(6\beta)}+\binom62\left(e^{i6\beta}\right)^{(6-2-2)}+\binom64(e^{i\beta})^{-2}\right\}$$

$$=2\cos6\beta+2\cdot15\cos2\beta$$

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$$\cos^{6} (x)-\sin^{6}(x)=(\cos^{2}(x)-\sin^{2}(x))(\cos^{4} (x)+\cos^{2}(x)\sin^{2}(x)+\sin^{4}(x))$$ $$a^{2}+b^{2}=(a+b)-2ab.$$ $$=\cos (2 x)(1-\cos^{2} (x)\sin^{2}(x))$$ $$=\cos 2x-\dfrac{1}{2}\cos (2x)\sin^{2}(2x)$$ $$=\dfrac{\cos 2x(1+\cos^{2}(2x))}{2}....؟$$

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Use $$\cos2A=2\cos^2A-1=1-2\sin^2A$$ to find

$$8(\cos^6 \beta - \sin^6 \beta)=(2\cos^2\beta)^3-(2\sin^2\beta)^3=(1+\cos2\beta)^3-(1-\cos2\beta)^3=?$$

Finally use $$\cos3B=4\cos^3B-3\cos B\iff4\cos^3B=?$$

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