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How to prove the following equation by a quick method? \begin{eqnarray} \\\sin ^6x+\cos ^6x=\frac{1}{8}\left(3\cos 4x+5\right)\\ \end{eqnarray}

If I use so much time to expand it and take extra care of the calculation process, I can find the answer. Since it is so easy to get mistakes, I have to try it 3 time to obtain the correct answer. Therefore, I wonder there are other easy ways to deal with this question. Is it right?

Thank you for your attention

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    $\begingroup$ $x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)$ seems like a first obvious step. $\endgroup$ – Thomas Andrews Oct 28 '13 at 16:52
  • $\begingroup$ @Thomas Andrews, I only know how to use Binomial Expansion. Would you mind giving me the reference for learning the expansion of $x^n+y^n$? Thank you. $\endgroup$ – Casper Oct 28 '13 at 17:02
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Using $a^3+b^3=(a+b)^3-3ab(a+b),$

$$(\sin^2x)^3+(\cos^2x)^3=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)$$

$$=1-3\frac{(2\sin x\cos x)^2}4$$

$$=1-\frac34(\sin^22x)\text{ using }\sin2x=2\sin x\cos x$$

$$=1-\frac34\frac{(1-\cos4x)}2 \text{ using }\cos2y=1-2\sin^2y$$

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  • $\begingroup$ @BabakS., thanks. I have rectified one mistake. $\endgroup$ – lab bhattacharjee Oct 28 '13 at 16:57
  • $\begingroup$ Nice Answer~! It is so quick. $\endgroup$ – Casper Oct 28 '13 at 17:38
  • $\begingroup$ @CasperLi, Thanks. hope I could convey the pattern $\endgroup$ – lab bhattacharjee Oct 28 '13 at 17:40
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To flesh out Thomas's comment into another answer: start with the factorization $x^3-y^3 = (x-y)(x^2+xy+y^2)$; this can be derived easily from the finite geometric series. Replacing $y$ with $-y$ yields $x^3+y^3 = (x+y)(x^2-xy+y^2)$, and then replacing $x$ with $x^2$ and $y$ with $y^2$ yields $x^6+y^6 = (x^2+y^2)(x^4-x^2y^2+y^4)$. Now, plugging in $x=\cos\theta, y=\sin\theta$ and using $\cos^2\theta+\sin^2\theta=1$ yields $\cos^6\theta+\sin^6\theta = \cos^4\theta-\cos^2\theta\sin^2\theta+\sin^4\theta$. From here, the simplest route is through an approach similar to lab bhattacharjee's answer: $\cos^4\theta-\cos^2\theta\sin^2\theta+\sin^4\theta$ $=(\cos^4\theta+2\cos^2\theta\sin^2\theta+\sin^4\theta)-3\cos^2\theta\sin^2\theta$ $=(\cos^2\theta+\sin^2\theta)^2-3\cos^2\theta\sin^2\theta$ $=1-3\cos^2\theta\sin^2\theta$; and now things proceed exactly as in that answer, using the double-angle formulas for sin and cos in turn.

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Using $\cos2A=1-2\sin^2A=2\cos^2A-1,$

$$(2\sin^2x)^3+(2\cos^2x)^3=(1-\cos2x)^3+(1+\cos2x)^3=2[1+3\cos^22x]$$

Now use $2\cos^22x-1=\cos2(2x)$

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