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In this question, the only proof of the trigonometric identity:

$$\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$$

is via factoring the sum of cubes: $$\cos^6{\theta}+\sin^6{\theta}=(\cos^2{\theta}+\sin^2{\theta})(\cos^4{\theta}-\cos^2{\theta}\sin^2{\theta}+\sin^4{\theta})$$

Is there another way to prove this identity? I'm hoping it will give more insight on it.

EDIT: Sorry for not clarifying before. I have learned everything up to and including first-semester calculus.

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    $\begingroup$ There are several ways to do so, but it will help us to know what your mathematical background is. What is your experience level with complex variables? $\endgroup$ – Cameron Buie Jun 13 '15 at 17:49
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    $\begingroup$ This is fine, but one side it's $x$ and other side $\theta$ ? $\endgroup$ – Mann Jun 13 '15 at 17:49
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    $\begingroup$ Thanks for pointing it out, I just fixed it. $\endgroup$ – Alnus Jun 13 '15 at 17:51
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$$s_1^6+c_1^6=\left(\frac{1-c_2}2\right)^3+\left(\frac{1+c_2}2\right)^3=\frac{2+6c_2^2}8=\frac{5+3c_4}8.$$

(The subscript obviously denotes the coefficient in front of $\theta$.)


The transformation

$$(s_1^6+c_1^6)=(s_1^2+c_1^2)(s_1^4-s_1^2c_1^2+c_1^4)=(s_1^2+c_1^2)^2-3s_1^2c_1^2=1-\frac34s_2^2$$ is also handy.

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  • $\begingroup$ May want to add for clarity that you used $2\cos^(2t)=\cos(4t)+1$ in the last step. $\endgroup$ – John Molokach Jun 13 '15 at 19:15
  • $\begingroup$ It won't fit anymore on a single line :( , and I used the same substitution just before. $\endgroup$ – Yves Daoust Jun 13 '15 at 19:16
  • $\begingroup$ Dauost, IMHO, this is BTW the most elegant solution by far. $\endgroup$ – John Molokach Jun 13 '15 at 19:17
  • $\begingroup$ Yep, the trick is to exploit the factorization of $6$ and switch to $2t$ as soon as possible. Thanks. $\endgroup$ – Yves Daoust Jun 13 '15 at 19:30
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A slightly different way, if you know the expansion $(a+b)^3=a^3+3a^2b+3ab^2+b^3$, is to use it with $a=\cos^2\theta$ and $b=\sin^2\theta$. In simplifying, you need also to factorize and use the identities $\cos^2\theta+\sin^2\theta=1$, $\sin2\theta=2\cos\theta\sin\theta$, and $\cos4\theta=1-2\sin^22\theta$. A similar way will work also for higher powers, for example $\cos^8\theta+\sin^8\theta$.

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If you use $\cos^2\theta = 1-\sin^2\theta$ (and vice versa), you have $$ \cos^6\theta+\sin^6\theta=(1-\sin^2\theta)\cos^4\theta+(1-\cos^2\theta)\sin^4\theta\\ =\cos^4\theta-\sin^2\theta\cos^4\theta-\cos^2\theta\sin^4\theta+\sin^4\theta\\ =\cos^4\theta-(\sin^2\theta+\cos^2\theta)\sin^2\theta\cos^2\theta+\sin^4\theta\\ =\cos^4\theta-\sin^2\theta\cos^2\theta+\sin^4\theta... $$ here you only had to see that the two center terms had a common factor, which may be simpler than factoring the cubic at one go.

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You can use your calculus knowledge:$$d/d\theta (cos^6\theta +sin^6\theta -1/8(5+3cos4\theta)) \\ =-6cos^5\theta sin\theta +6sin^5\theta cos\theta +3/2sin4\theta \\ =6sin\theta cos\theta (sin^4\theta -cos^4\theta)+3sin2\theta cos2\theta \\ = -3sin2\theta cos2\theta +3sin2\theta cos2\theta =0 \\ \Rightarrow cos^6\theta +sin^6\theta -1/8(5+3cos4\theta )=C $$ for some constant $C$. $\\ \theta =0$ gives $C=0$. Thus, $cos^6\theta +sin^6\theta =1/8(5+3cos4\theta )$.

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I don't know if your course covered complex numbers. If so, you can use the Euler's identity to boil this down to a simple polynomial expression.

$$\cos\theta = \frac12(e^{i\theta}+e^{-i\theta})$$ $$\sin\theta = \frac1{2i}(e^{i\theta}-e^{-i\theta})$$ Let's just say $z=e^{i\theta}$ to shorten the notation. You have

$$\frac{1}{2^6}(z+1/z)^6-\frac{1}{2^6}(z-1/z)^6=$$ $$\frac{1}{2^6}\left((z^6+6z^4+15z^2+20+15z^{-2}+6z^{-4}+z^{-6})-(z^6-6z^4+15z^2-20+15z^{-2}-6z^{-4}+z^{-6})\right)=$$ $$=\frac{1}{64}\left(40+24\cdot\frac12(z^4+z^{-4})\right)$$ $$=\frac{5}{8}+\frac{3}{8}\cos4\theta$$

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I have two methods:

Using Calculus: Define $f(\theta)=\sin{\theta}\cos^5{\theta}-\sin^5{\theta}\cos{\theta}$. Now, we calculate:

$$\int{(f(\theta)\mathrm{d}\theta)}$$

in two ways. First, calculate it directly (using $u$-subs):

$$\int{(f(\theta)\mathrm{d}\theta)}=\int{((\sin{\theta}\cos^5{\theta}-\sin^5{\theta}\cos{\theta})\mathrm{d}\theta)}=-\frac{1}{6}(\cos^6{\theta}+\sin^6{\theta})+C_1$$

Second, simplify $f(\theta)$ first, and integrate afterwards:

$$f(\theta)=\sin{\theta}\cos^5{\theta}-\sin^5{\theta}\cos{\theta}=\sin{\theta}\cos{\theta}(\sin^4{\theta}-\cos^4{\theta})=\sin{\theta}\cos{\theta}(\sin^2{\theta}-\cos^2{\theta})(\sin^2{\theta}+\cos^2{\theta})=\sin{\theta}\cos{\theta}(\sin^2{\theta}-\cos^2{\theta})=\frac{1}{2}(2\sin{\theta}\cos{\theta})(-(\cos^2{\theta}-\sin^2{\theta}))=-\frac{1}{2}(\sin{2\theta})(\cos{2\theta})=-\frac{1}{4}(2\sin{2\theta}\cos{2\theta})=-\frac{1}{4}\sin{4\theta}$$

Now, we integrate $f(\theta)$:

$$\int{(f(\theta)\mathrm{d}\theta)}=\int{(-\frac{1}{4}\sin{4\theta}\mathrm{d}\theta)}=-\frac{1}{16}\cos{4\theta}+C_2$$

Of course, the point is that the two expressions are equal! i.e.:

$$-\frac{1}{6}(\cos^6{\theta}+\sin^6{\theta})+C_1=-\frac{1}{16}\cos{4\theta}+C_2$$

If we let $C_3$ be a single constant (instead of $C_1$ and $C_2$), and simplify:

$$\cos^6{\theta}+\sin^6{\theta}+C_3=\frac{3}{8}\cos{4\theta}$$

Plugging in a convenient value such as $\theta=0$ to find $C_3$:

$$\cos^6{0}+\sin^6{0}+C_3=\frac{3}{8}\cos{4(0)}$$

$$C_3=-\frac{5}{8}$$

Using the value of $C_3$:

$$\cos^6{\theta}+\sin^6{\theta}-\frac{5}{8}=\frac{3}{8}\cos{4\theta}$$

$$\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$$

Using Algebra: A slightly different way using algebra. We know that:

$$(x+y)^3=x^3+3x^2y+3xy^2+y^3=x^3+y^3+3xy(x+y)$$

so:

$$1=1^3=(\sin^2{\theta}+\cos^2{\theta})^3=\sin^6{\theta}+\cos^6{\theta}+3\sin^2{\theta}\cos^2{\theta}(\sin^2{\theta}+\cos^2{\theta})=\sin^6{\theta}+\cos^6{\theta}+3\sin^2{\theta}\cos^2{\theta}$$

$$1=\sin^6{\theta}+\cos^6{\theta}+3\sin^2{\theta}\cos^2{\theta}$$

$$\frac{3}{8}+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}+3\sin^2{\theta}\cos^2{\theta}$$

$$\frac{3}{8}-3\sin^2{\theta}\cos^2{\theta}+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$

$$\frac{3}{8}-\frac{3}{4}(4\sin^2{\theta}\cos^2{\theta})+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$

$$\frac{3}{8}-\frac{3}{4}(2\sin{\theta}\cos{\theta})^2+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$

$$\frac{3}{8}-\frac{3}{4}\sin^2{2\theta}+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$

$$\frac{3}{8}(1-2\sin^2{2\theta})+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$

$$\frac{3}{8}(\cos{4\theta})+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$

$$\frac{1}{8}(5+3\cos{4\theta})=\sin^6{\theta}+\cos^6{\theta}$$

I hope you enjoy these two very elegant proofs!

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  • $\begingroup$ You can do much more elegant by factoring $(\sin^2\theta)^3+(\cos^2\theta)^3$. $\endgroup$ – Yves Daoust Nov 29 '18 at 8:12

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