If sin theta + cos theta = x, then {sin ^6}th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $\sin 	heta + \cos 	heta = x.$Squaring both sides,we get $\sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = x^2.$Since $\sin^2 \

Vedclass · Accepted Answer

${x^2} \le 2$

Vedclass · Answer

(B) Given $\sin 	heta + \cos 	heta = x.$Squaring both sides,we get $\sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = x^2.$Since $\sin^2 	heta + \cos^2 	heta = 1,$ we have $1 + \sin 2	heta = x^2,$ which implies $\sin 2	heta = x^2 - 1.$Since the range of $\sin 2	heta$ is $[-1, 1],$ we must have $-1 \le x^2 - 1 \le 1,$ which simplifies to $0 \le x^2 \le 2.$Now,consider the expression ${\sin ^6}	heta + {\cos ^6}	heta.$Using the identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b),$ where $a = \sin^2 	heta$ and $b = \cos^2 	heta,$${\sin ^6}	heta + {\cos ^6}	heta = (\sin^2 	heta + \cos^2 	heta)^3 - 3 \sin^2 	heta \cos^2 	heta (\sin^2 	heta + \cos^2 	heta).$$= 1^3 - 3(\sin 	heta \cos 	heta)^2 (1) = 1 - 3(\frac{\sin 2	heta}{2})^2 = 1 - \frac{3}{4} \sin^2 2	heta.$Substituting $\sin 2	heta = x^2 - 1,$$= 1 - \frac{3}{4}(x^2 - 1)^2 = \frac{1}{4}[4 - 3(x^2 - 1)^2].$This identity holds true only when $\sin 2	heta$ is defined,which requires $x^2 \le 2.$

If $\sin \theta + \cos \theta = x,$ then ${\sin ^6}\theta + {\cos ^6}\theta = \frac{1}{4}[4 - 3{({x^2} - 1)^2}]$ for

Similar Questions

If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n,$ then ${a^2} + {b^2} = $

If $x = a \sin \theta - b \cos \theta$ and $y = a \cos \theta + b \sin \theta$,then which of the following is true?

The value of $\cos \frac{\pi}{10} \cos \frac{2\pi}{10} \cos \frac{4\pi}{10} \cos \frac{8\pi}{10} \cos \frac{16\pi}{10}$ is:

If $A + B + C = 180^o$,then the value of $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}$ is equal to

If $\theta+\phi=\frac{2 \pi}{3}$ and $\cos \theta=\frac{\sqrt{3}}{2},$ what is the value of $\sin \phi ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module