Question

sin^4x/3+cos^4x/4=1/7 then the value of sin^8x/27+cos^8x/64 is equal to

Answer 1

Answer-

$\boxed { \boxed { \bolded {\dfrac{\sin^8x}{27}+\dfrac{\cos^8x}{64}= \frac{4}{2401}}}}$

Solution-

Given,

$\Rightarrow \dfrac{\sin^4x}{3}+\dfrac{\cos^4x}{4}=\dfrac{1}{7}$

$\Rightarrow 4\sin^4x+3\cos^4x=\dfrac{12}{7}$

$\Rightarrow 4\sin^4x+3(\cos^2x)^2=\dfrac{12}{7}$

$\Rightarrow 4\sin^4x+3(1-\sin^2x)^2=\dfrac{12}{7}$

$\Rightarrow 4\sin^4x+3(1+\sin^4x-2\sin^2x)=\dfrac{12}{7}$

$\Rightarrow 4\sin^4x+3+3\sin^4x-6\sin^2x=\dfrac{12}{7}$

$\Rightarrow 7\sin^4x-6\sin^2x+3=\dfrac{12}{7}$

$\Rightarrow 49\sin^4x-42\sin^2x+21=12$

$\Rightarrow 49\sin^4x-42\sin^2x+9=0$

$\Rightarrow 49(\sin^2x)^2-42(\sin^2x)+9=0$

Solving the equation,

$\sin^2x=\dfrac{42\pm\sqrt{42^2-4\times49\times 9}}{2\times 49}$

$=\dfrac{42\pm\sqrt{0}}{2\times 49}$

$=\dfrac{42}{2\times 49}=\dfrac{3}{7}$

Therefore,

$=\dfrac{\sin^8x}{27}+\dfrac{\cos^8x}{64}$

$=\dfrac{(\sin^2x)^4}{27}+\dfrac{(\cos^2x)^4}{64}$

$=\dfrac{(\sin^2x)^4}{27}+\dfrac{(1-\sin^2x)^4}{64}$

$=\dfrac{(\frac{3}{7})^4}{27}+\dfrac{(1-\frac{3}{7})^4}{64}$

$=\dfrac{(\frac{3}{7})^4}{27}+\dfrac{(\frac{4}{7})^4}{64}$

$=\frac{1}{27}{(\frac{3}{7})^4}+\frac{1}{64}{(\frac{4}{7})^4}$

$=\frac{1}{27}{(\frac{81}{2401})}+\frac{1}{64}{(\frac{64}{2401})}$

$=\frac{3}{2401}+\frac{1}{2401}$

$=\frac{4}{2401}$

$\boxed { \boxed { \bolded { \therefore \dfrac{\sin^8x}{27}+\dfrac{\cos^8x}{64}= \frac{4}{2401}}}}$