MHB Proving Identity 46: $\cos^{6}(A)+\sin^{6}(A)=1-3\sin^{2}(A)\cos^{2}(A)$

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The identity $\cos^{6}(A)+\sin^{6}(A)=1-3\sin^{2}(A)\cos^{2}(A)$ is being verified as part of a series of 53 trigonometric identities. The proof involves manipulating the left-hand side using the identity $\sin^2(A) + \cos^2(A) = 1$ and expanding the terms. The calculations show that the left side simplifies to $1 - 3\sin^2(A)\cos^2(A)$, confirming the identity. This verification process highlights the relationship between powers of sine and cosine in trigonometric identities. The discussion emphasizes the importance of algebraic manipulation in proving such identities.
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Again I'm stuck with another problem in proving identities. This is the 46th item in the list of 53 identities that I'm asked to verify and so far I was able to prove 45 of them, there are 8 items left and one of them is this Identity

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

Thanks!
 
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Hello, paulmdrdo!

$\cos^6\!A+\sin^6\!A \;=\;1-3\sin^2\!A\cos^2\!A$
$\begin{array}{ccc}\sin^6\!A + \cos^6\!A &=& \underbrace{(\sin^2\! A+\cos^2\! A)}_{\text{This is 1}}(\sin^4\!A - \sin^2\!A\cos^2\!A + \cos^4\!A) \\ &=& (\sin^4\!A + 2\sin^2\!A\cos^2\!A + \cos^4\!A) - 3\sin^2\!A\cos^2\!A \\ \\ &=& (\sin^2\!A + \cos^2\!A)^2 - 3\sin^2\!A\cos^2\!A \\ \\ &=& 1 - 3\sin^2\!A\cos^2\!A \end{array}$
 
[math]\begin{align*}\cos^6(A)+\sin^6(A)&=(1-\sin^2(A))^3+(1-\cos^2(A))^3 \\&=1-3\sin^2(A)+3\sin^4(A)-\sin^6(A)+1-3\cos^2(A)+3\cos^4(A)-\cos^6(A) \\2(\cos^6(A)+\sin^6(A))&=-1+3(\cos^4(A)+\sin^4(A)) \\&=-1+3[(\cos^2(A)+\sin^2(A))^2-2\cos^2(A)\sin^2(A)] \\&=-1+3-6\cos^2(A)\sin^2(A) \\\cos^6(A)+\sin^6(A)&=1-3\cos^2(A)\sin^2(A)\end{align*}[/math]
 
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