Sine and Cosine powers

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

$\left(\sin^2\theta\right)^{3}+\left(\cos^{2}\theta\right)^{3}=2\sin\theta\cos\theta$

$\left(\sin^{2}\theta+\cos^{2}\theta\right)\left(\sin^{4}\theta-\sin^{2}\theta\cos^{2}\theta+\cos^{4}\theta\right)=2\sin\theta\cos\theta$

$\left(\left(\sin^{2}\theta\right)^{2}+\left(\cos^{2}\theta\right)^{2}\right)-sin^{2}\theta\cos^{2}\theta=2\sin\theta\cos\theta$

$\left(\sin^{2}\theta+\cos^{2}\theta\right)^{2}-2\sin^{2}\theta\cos^{2}\theta-\sin^{2}\theta\cos^{2}\theta=2\sin\theta\cos\theta$

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

Let $\sin\theta\cos\theta=x$ .

$1-3x^{2}=2x$

$3x^{2}+2x-1=0$

$(3x-1)(x+1)=0$

$x=\frac{1}{3}~\text{or}~x=-1$

Now, $x\neq-1$ . (Why?)

$\therefore \sin\theta\cos\theta=\frac{1}{3}\Rightarrow\boxed{\sin(2\theta)=\frac{2}{3}}$

Got the idea from this !

I used the identity $\displaystyle \sin^{6} x + \cos^{6} x = 1-\frac{3}{4}\sin^{2} 2x$ . Then, all I had to do was $\displaystyle1-\frac{3}{4}\sin^{2} 2x = \sin2x$ and solve a quadratic equation, substituting $y$ for $\sin 2x$ and solving for $y$ . The two results were $\displaystyle\frac{2}{3}$ and $-2$ , but only $\displaystyle\frac{2}{3}$ can be a response because $\arcsin ( -2)$ is not defined.

(sin^2(P))^3+(cos^2(P))^3=sin2(p). cos2(p)=1-2sin^2(p)=2cos^2(p)-1. so (.5-.5cos2(p))^3+(.5+.5cos2(p))^3=sin2p. (1-cos2p)^3+(1+cos2p)^3=8sin2p. (1-cos2p)×(1-2cos2p+cos^2(2p))+(1+cos2p)×(1+2cos2p+cos^2(2p))=8sin2p. so the above equation will be shortcut as 2+6cos^2(2p)=8sin2p =1+3×(1-sin^2(2p))=. 4sin2p =4-3sin^2(2p) put sin2p=y. 3y^2+4y-4=0 so y=-2 (rejected) or 2÷3 So y=sin2p=2÷3#######

( (sin a)^2 + (cos a)^2 )^3 = (sin a)^6 + (cos a)^6 + 3 x (sin a)^4 x (cos a)^2 + 3 x (sin a)^2 x cos a)^4

1 = (sin a)^6 + (cos a)^6 + 3 x (sin a)^2 x (cos a)^2 x ( (sin a)^2 + (cos a)^2 )

1 = sin 2a + (3/4) x (sin 2a)^2

Solving with the condition |sin 2a| <= 1 yields sin 2a = 2/3.

$\begin{aligned} \color{#3D99F6}{\sin^6 \theta + \cos^6 \theta} & = \sin (2\theta) \quad \quad \small \color{#3D99F6}{\text{Since }\cos (2\theta) = 1-2\sin^2 \theta = 2\cos^2 \theta -1} \\ \color{#3D99F6}{\frac{1}{2^3} (1-\cos (2 \theta))^3 +\frac{1}{2^3} (1-\cos (2 \theta))^3 } & = \sin (2 \theta) \\ \frac{2}{8}(\color{#3D99F6}{1 + 3 \cos^2 (2 \theta)}) & = \sin (2 \theta) \quad \quad \small \color{#3D99F6}{\text{Even terms cancel out.}} \\ 1 + 3(1- \sin^2 (2 \theta)) & = 4 \sin (2 \theta) \\ \Rightarrow 3 \sin^2 (2 \theta) + 4 \sin (2 \theta) - 4 & = 0 \\ (3 \sin (2 \theta) - 2)(\sin (2 \theta) + 2) & = 0 \\ \Rightarrow \sin (2 \theta) & = \boxed {\frac{2}{3}} \quad \quad \small \color{#3D99F6}{\sin (2 \theta) \ne - 2} \end{aligned}$

Same but ending a little differently. $\sin^{6}\theta+\cos^{6}\theta=\sin(2\theta)$

$\left(\sin^2\theta\right)^{3}+\left(\cos^{2}\theta\right)^{3}=\sin(2\theta)$

$\left(\sin^{2}\theta+\cos^{2}\theta\right)\left(\sin^{4}\theta-\sin^{2}\theta\cos^{2}\theta+\cos^{4}\theta\right)=\sin(2\theta)$

$\left(\left(\sin^{2}\theta\right)^{2}+\left(\cos^{2}\theta\right)^{2}\right)-\dfrac{sin(2\theta)}{4}=\sin(2\theta)$

$\left(\sin^{2}\theta+\cos^{2}\theta\right)^{2}-\dfrac{3*sin(2\theta)}{4}=\sin(2\theta)$

$1-\dfrac{3*\sin(2\theta)}{4}=\sin(2\theta) \\ 4-3*\sin(2\theta)^2-4*\sin(2\theta)=0\\$ $\sin(2\theta)=\dfrac{4\pm \sqrt{4^2+4*4*4} } {2(-3)}=~~~~\color{#D61F06}{\dfrac 2 3}~~~~~~~~~~~-2 ~not~ a~ solution.$