If A = sin^8 theta + cos^{14} theta,then for | vedclass

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

Question

(B) Given that $A = \sin^8 	heta + \cos^{14} 	heta$.Since $0 \le \sin^2 	heta \le 1$ and $0 \le \cos^2 	heta \le 1$ for all real $	heta$,we have:

Vedclass · Accepted Answer

$0 < A \le 1$

Vedclass · Answer

(B) Given that $A = \sin^8 	heta + \cos^{14} 	heta$.Since $0 \le \sin^2 	heta \le 1$ and $0 \le \cos^2 	heta \le 1$ for all real $	heta$,we have:$\sin^8 	heta \le \sin^2 	heta$ and $\cos^{14} 	heta \le \cos^2 	heta$.Adding these inequalities,we get:$A = \sin^8 	heta + \cos^{14} 	heta \le \sin^2 	heta + \cos^2 	heta = 1$.Also,since $\sin^8 	heta > 0$ and $\cos^{14} 	heta > 0$ for all $	heta$ (excluding cases where one is zero,but they cannot both be zero simultaneously),$A > 0$.Therefore,$0 < A \le 1$.

If $A = \sin^8 \theta + \cos^{14} \theta$,then for all real values of $\theta$:

Similar Questions

$\frac{1}{4} [\sqrt{3} \cos 23^\circ - \sin 23^\circ] = $

If $\sin \theta = -\frac{7}{25}$ and $\theta$ is in the third quadrant,then $\frac{7 \cot \theta - 24 \tan \theta}{7 \cot \theta + 24 \tan \theta} = $

$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = $

If $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b},$ then $\frac{\tan x}{\tan y}$ is equal to

The value of $\cot(7.5^{\circ}) + \tan(67.5^{\circ}) - \cot(67.5^{\circ}) - \tan(7.5^{\circ})$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module