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I have given the task to find $\sin^6(a)+\sin^6(b)+\sin^6(c)$. It is also given $a , b$ and $c$ are real .

Further given down is only my attempt , its your choice you want to read it not . I just want a solution which could be completing my solution or give a better and simpler method . Any help like another attempt , links , hints etc are appreciated. Solution can include use of complex numbers or trigonometry. It would be appreciated if you don't use tools higher than high school level .

First thing I noticed is he above equation can be written as $[\sin(a)+\sin(b)+\sin(c)]^2\ +\ [\cos(a)+\cos(b)+\cos(c)]^2\ =\ 0$

Which implies $\cos(a)+\cos(b)+\cos(c)=0$ .....(1) and also $\sin(a)+\sin(b)+\sin(c)=0$ ....(2) .

$NA =\cos(Na)+ i\sin(Na)$

$NB =\cos(Nb)+ i\sin(Nb)$

$NC =\cos(Nc)+ i\sin(Nc)$

Where $N$ is an arbitrary natural number and $i$ is the square root of $-1$.

From (1) and (2)

$A+B+C=0 .. (3)$

Which implies $\frac{1}{AB} + \frac{1}{BC} + \frac{1}{BC}=0$

Taking conjugate both sides $AB+BC+CA=0$ ....(4)

So $A²+B²+C²=0$ .....(5) also as $(A+B+C)^2=0$

We can also by repeating same procedure $A^4+B^4+C^4=0$ ....(6)

Now $\cos²(a)+\cos²(b)+\cos²(c)$ = $\frac{1}{2}[3-\cos(2a)-\cos(2b)-\cos(2c)]$ = $\frac{3}{2}$

As by (5) we know $\cos(2a)+\cos(2b)+\cos(2c)=0$

Same way $\sin^2(a)+\sin^2(b)+\sin^2(c)= \frac{3}{2}$

Also $\cos^2(2a)+\cos^2(2b)+\cos^2(2c)=\frac{3}{2}$ as $\cos(4a) + \cos(4b) + \cos(4c) = 0 $ by (6)

As $[\cos(a) + \cos(b) + \cos(c)]^2 = 0 $, so $\cos(a)\cos(b)+ \cos(b)\cos(c)+\cos(c)\cos(a)= \frac{-3}{4} $

Same way $\cos(2a)\cos(2b)+\cos(2b)\cos(2c) +\cos(2c)\cos(2a)= \frac{-3}{4} $

$[\sin^2(a)-\sin^2(b)]^2+ [\sin^2(b)-\sin^2(c)]^2+ [\sin^2(c)-\sin^2(a)]^2$

= $\frac{1}{4} \left([\cos(2a)-\cos(2b)]^2+ [\cos(2b)-\cos(2c)]^2 +[\cos(2c)-\cos(2a)]^2\right)$

= $\frac{9}{8}$

$l³+m³+n³=(l+m+n)(l²+m²+n²-lm-mn-nm) + 3lmn $

$l=\sin²(a),\:m=\sin²(b)$ and $n=\sin²(c)$

I have already found $l+m+n$ and $l²+m²+n²-lm-mn-nm$ above. So I am stuck with finding $lmn$.

Please help me find $lmn$ or provide another easier solution as you can see this method is really lenghty and tedious . Please also make out any mistakes I have made .

Bernard
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aryan bansal
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  • See https://math.stackexchange.com/questions/639890/minimum-value-of-cosa-b-cosb-c-cosc-a-is-3-2 and https://math.stackexchange.com/questions/1382047/trigonometry-problem and https://math.stackexchange.com/questions/1397066/clarification-regarding-a-question – lab bhattacharjee Jul 16 '20 at 12:32
  • @lab bhattacharjee thank you for your effort but these dont help me though – aryan bansal Jul 16 '20 at 13:31

1 Answers1

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From your notes: $$\cos a+\cos b+\cos c=0 \Rightarrow \cos^2a+\cos^2b+2\cos a\cos b=\cos^2c\\ \sin a+\sin b+\sin c=0\Rightarrow \sin^2a+\sin^2b+2\sin a\sin b=\sin^2c\\ $$ Adding the two: $$\cos(a-b)=-\frac12.$$ Similarly: $$\cos(b-c)=-\frac12\\ \cos(c-a)=-\frac12$$ One solution is: $a=180^\circ,b=60^\circ,c=-60^\circ$. Hence, the sum of sixth powers of sines is $0$.

farruhota
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