Not Six Times its Angle

another easy solution: $\sin ^{ 6 }{ x } +\cos ^{ 6 }{ x } ={ \left( \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } \right) }^{ 3 }-3\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } \left( \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } \right) \\ \quad \quad \quad =1-\frac { 3 }{ 4 } \sin ^{ 2 }{ 2x } \times 1$ .On substituting values we get $=1-\frac { 1 }{ 12 } =\frac { 11 }{ 12 }$

Noting that $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$ , $\sin^{2}(\theta) + \cos^{2}(\theta) = 1$ and

$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ we find that

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

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

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

$1 - 3*\dfrac{\sin^{2}(2\theta)}{4} =1 - (\dfrac{3}{4})(\dfrac{1}{3})^{2} = 1 - \dfrac{1}{12} = \dfrac{11}{12}.$

Thus $a + b = 11 + 12 = \boxed{23}.$

$\sin{2\theta} = \dfrac {1}{3}\quad \Rightarrow \cos{2\theta} = \dfrac {2\sqrt{2}} {3} \quad \Rightarrow 2\cos^2 {\theta} - 1 = \dfrac {2\sqrt{2}} {3}$

$\Rightarrow \cos^2 {\theta} = \frac {1}{2} + \frac {\sqrt{2}}{3}\quad \Rightarrow \sin^2 {\theta} = \frac {1}{2} - \frac {\sqrt{2}}{3}$

$\sin^6 {\theta} + \cos^2 {\theta} = \left( \frac {1}{2} - \frac {\sqrt{2}}{3} \right)^3 + \left( \frac {1}{2} + \frac {\sqrt{2}}{3} \right)^3$

$= \left( \frac {1}{8} - 3 (\frac{1}{4})(\frac {\sqrt{2}}{3}) + 3 (\frac{1}{2})(\frac {2}{9}) - \frac {2\sqrt{2}}{27} \right) + \left( \frac {1}{8} + 3 (\frac{1}{4})(\frac {\sqrt{2}}{3}) + 3 (\frac{1}{2})(\frac {2}{9}) + \frac {2\sqrt{2}}{27} \right)$

$=2\left( \frac {1}{8} + \frac {1}{3} \right) = \frac {11}{12}$

$\Rightarrow a + b = \boxed{23}$

Factoring the sum of cubes:

$sin^{6}\theta + cos^{6}\theta = (sin^{2}\theta + cos^{2}\theta)(sin^{4}\theta - sin^{2}\theta *cos^{2}\theta + cos^{4}\theta)$

$= sin^{4}\theta - sin^{2}\theta *cos^{2}\theta + cos^{4}\theta$

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

$= (cos^{2} 2\theta) + \frac{1}{4}sin^{2} 2\theta$

$= (1 - sin^{2} 2\theta) + \frac{1}{4}sin^{2} 2\theta$

$= (1 - \frac{1}{9}) + \frac{1}{36}$

$= \frac{8}{9} + \frac{1}{36}$

$= \frac{32 + 1}{36}$

$= \frac{33}{36} = \frac{11}{12}$

Thus, $a = 11$ , $b = 12$ , $a + b = 23$

Same with mudit

$(sin^2 \theta + cos^2 \theta )^3 = sin^6 \theta + cos^6 \theta + 3(sin^2 \theta cos^2\theta)(sin^2 \theta+ cos^2 \theta)$ Using sin2 $\theta$ = 2sin $\theta$ cos $\theta$ $\Rightarrow\sin^6 \theta + cos^6 \theta = 1 - \frac{3}{4}sin^{2} 2\theta = 1 - \frac{1}{12} = \boxed{\frac{11}{12}}$

$sin\quad 2\theta \quad =\quad 2.sin\theta .cos\theta \quad =\quad \frac { 1 }{ 3 } .........(1)\\ sin^{ 2 }\theta \quad +\quad cos^{ 2 }\theta \quad =\quad 1\quad ................(2)\\ \\ sin^{ 6 }\theta \quad +\quad cos^{ 6 }\theta \quad \\ \Rightarrow \quad (sin^{ 2 }\theta +cos^{ 2 }\theta )^{ 3 }\quad -\quad 3\quad sin^{ 2 }\theta cos^{ 2 }\theta \quad (sin^{ 2 }\theta +cos^{ 2 }\theta )\\ \\ Putting\quad the\quad values\quad from\quad (1)\quad and\quad (2)\\ \Rightarrow \quad { 1 }^{ 3 }\quad -\quad 3.{ \left\{ \frac { 1 }{ 6 } \right\} }^{ 2 }.(1)\\ \Rightarrow 1\quad -\quad \frac { 1 }{ 12 } \\ \Rightarrow \frac { 11 }{ 12 } \\ Let\quad \frac { 11 }{ 12 } \quad be\quad represented\quad as\quad \frac { a }{ b } \\ \\ Finally\quad ,\quad The\quad answer\quad will\quad be\quad \boxed { a+b } \quad i.e.\quad \boxed { 23 }$

sin2A= 2sinAcosA

→ sinAcosA= 1/6

Also sin^6A + cos^6A = (sin^2A + cos^2A)^3 - 3sin^2Acos^2 A( sin^2A + cos^2A)

substitute value of sinAcosA nget d ans.