How Can You Prove the Trigonometric Identity Cos^6A+Sin^6A=1-3Sin^2ACos^2A?

[Silver Bolt]

Prove $Cos^6A+Sin^6A=1-3 \hspace{0.2cm}Sin^2 A\hspace{0.02cm}Cos^2A$

So far,

$Cos^6A+Sin^6A=1-3 \hspace{0.2cm}Sin^2 A\hspace{0.02cm}Cos^2A$
$L.H.S=(Cos^2A)^3+(Sin^2A)^3$
$=(Cos^2A+Sin^2A)(Cos^4A-Cos^2ASin^2A+Sin^4A)$
$=\underbrace{(Cos^2A+Sin^2A)}_{\text{1}}(Cos^4A-Cos^2ASin^2A+Sin^4A) $
$=1(Cos^4A-Cos^2ASin^2A+Sin^4A)$

Can someone help from here?

 

[Greg]

$$\sin^4(A)-\sin^2(A)\cos^2(A)+\cos^4(A)=(\sin^2(A)+\cos^2(A))^2-3\sin^2(A)\cos^2(A)$$

 

[Silver Bolt]

$$\sin^4(A)-\sin^2(A)\cos^2(A)+\cos^4(A)=(\sin^2(A)+\cos^2(A))^2-3\sin^2(A)\cos^2(A)$$

A little explanation on how did that -3 come in $\sin^2(A)+\cos^2(A))^2-3\sin^2(A)\cos^2(A)$

 

[Greg]

Expand $(\sin^2(A)+\cos^2(A))^2$. The result should answer your question.

 

[Silver Bolt]

$\sin^4(A)-\sin^2(A)\cos^2(A)+\cos^4(A)=(\sin^2(A)+\cos^2(A))^2-3\sin^2(A)\cos^2(A)$

Expand $(\sin^2(A)+\cos^2(A))^2$. The result should answer your question.

This part can be rewritten as

$\sin^4(A)+\cos^4(A)-\sin^2(A)\cos^2(A)=(\sin^2(A)+\cos^2(A))^2-\sin^2(A)\cos^2(A)$

Expanding $(\sin^2(A)+\cos^2(A))^2=\sin^4(A)+2\sin^2(A)\cos^2(A)+cos^4(A)$

Now using it instead of$ (\sin^2(A)+\cos^2(A))^2$

$\sin^4(A)+2\sin^2(A)\cos^2(A)+cos^4(A)-\sin^2(A)\cos^2(A)=\sin^4(A)+\sin^2(A)\cos^2(A)+cos^4(A)$

which is not the desired answer.Where have I gone wrong?

 

[Greg]

$\sin^4(A)-\sin^2(A)\cos^2(A)+\cos^4(A)=(\sin^2(A)+\cos^2(A))^2-3\sin^2(A)\cos^2(A)$

Start with the line above and expand the RHS.

 

[MarkFL]

An alternate approach would be to factor the LHS as the sum of two cubes:

$$\cos^6(A)+\sin^6(A)=\left(\cos^2(A)+\sin^2(A)\right)\left(\cos^4(A)-\cos^2(A)\sin^2(A)+\sin^4(A)\right)$$

Now, for the first factor on the RHS, we can apply a Pythagorean identity to state:

$$\cos^6(A)+\sin^6(A)=\cos^4(A)-\cos^2(A)\sin^2(A)+\sin^4(A)$$

We can also use this same Pythagorean identity to state:

$$1=\left(\cos^2(A)+\sin^2(A)\right)^2=\cos^4(A)+2\cos^2(A)\sin^2(A)+\sin^4(A)\implies \cos^4(A)+\sin^4(A)=1-2\cos^2(A)\sin^2(A)$$

And so, there results:

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