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Prove that ​$$6(\sin^{10}A+\cos^{10}A)-15(\sin^8A+\cos^8A)+10(\sin^6A+\cos^6A)-1=0​$$ Expression can be verified for different values of $A$ such as $\frac\pi4,\frac\pi2$ etc. But to prove it for general value of A?

Parcly Taxel
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H.P. Das
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2 Answers2

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Hint: Try to use $\cos^2 A = 1-\sin^2 A $

Writing $s$ for $\sin^2 A$ we have $6(s^5 + (1-s)^5) -15(s^4+ (1-s)^4) + 10(s^3 + (1-s)^3) - 1 = 6s^5 + 6 - 30s + 60s^2 - 60s^3 + 30s^4 - 6s^5 - 15s^4 - 15 + 60s - 90s^2 + 60s^3 - 15s^4 + 10s^3 + 10 - 30s + 30s^2 - 10s^3 - 1 = 0$

KoA
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  • Why the downvote? That approach works. – almagest Apr 22 '16 at 17:09
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    That is not very helpful since you will be applying the binomial theorem for ridiculous powers, thrice. But yes, you will be able to solve it. –  Apr 22 '16 at 17:11
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    Then show that it works..Provide at least a hint on why it works.. – MathematicianByMistake Apr 22 '16 at 17:11
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    Put $s=\sin^2 x$. The expression is $6s^5+6(1-s)^5-15s^4-15(1-s)^4+10s^3+10(1-s)^3$ $=6s^5+6-30s+60s^2-60s^3+30s^4-6s^5$ $-15s^4-15+60s-90s^2+60s^3-15s^4$ $+10s^3+10-30s+30s^2-10s^3$ $=(6-15+10)+s(-30+60-30)+s^2(60-90-30)$ $+s^3(-60+60+10-10)+s^4(30-15-15)+s^5(6-6)$ $=1$ – almagest Apr 22 '16 at 17:23
  • @MathematicianByMistake Editing takes a long time on the mobile app for me, sorry! – KoA Apr 22 '16 at 17:33
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Set $x=\sin^2A$ and $y=\cos^2A$. Then $x+y=1$ and \begin{align} 1&=(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\\ &=x^5+y^5+5xy(x^3+y^3)+10x^2y^2(x+y)\\ &=x^5+y^5+5xy(x+y)(x^2-xy+y^2)+10x^2y^2\\ &=x^5+y^5+5xy(x^2-xy+y^2)+10x^2y^2\\ &=x^5+y^5+5xy((x+y)^2-3xy)+10x^2y^2\\ &=x^5+y^5+5xy(1-3xy)+10x^2y^2\\ &=x^5+y^5-5x^2y^2+5xy \end{align} Similarly, \begin{align} 1&=(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4\\ &=x^4+y^4+4xy(x^2+y^2)+6x^2y^2\\ &=x^4+y^4+4xy(1-2xy)+6x^2y^2\\ &=x^4+y^4-2x^2y^2+4xy \end{align} and \begin{align} 1&=(x+y)^3=x^3+3x^2y+3xy^2+y^3\\ &=x^3+y^3+3xy(x+y)\\ &=x^3+y^3+3xy \end{align} Set $p=xy$; we have proved that \begin{align} x^5+y^5&=5p^2-5p+1\\ x^4+y^4&=2p^2-4p+1\\ x^3+y^3&=-3p+1 \end{align} so finally your expression is $$ 6(5p^2-5p+1)-15(2p^2-4p+1)+10(-3p+1)-1=0 $$

egreg
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