If A = sin^2 theta + cos^4 theta,then for all | vedclass

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

Question

(B) Given $A = \sin^2 	heta + \cos^4 	heta$.Since $\cos^2 	heta \le 1$,we have $\cos^4 	heta \le \cos^2 	heta$.Thus,$A = \sin^2 	heta + \cos^4 \

Vedclass · Accepted Answer

$\frac{3}{4} \le A \le 1$

Vedclass · Answer

(B) Given $A = \sin^2 	heta + \cos^4 	heta$.Since $\cos^2 	heta \le 1$,we have $\cos^4 	heta \le \cos^2 	heta$.Thus,$A = \sin^2 	heta + \cos^4 	heta \le \sin^2 	heta + \cos^2 	heta = 1$. So,$A \le 1$.Now,express $A$ in terms of $\cos^2 	heta$:$A = (1 - \cos^2 	heta) + \cos^4 	heta = \cos^4 	heta - \cos^2 	heta + 1$.Let $x = \cos^2 	heta$,where $0 \le x \le 1$.Then $A = x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$.Since $(x - \frac{1}{2})^2 \ge 0$,the minimum value is $\frac{3}{4}$ (at $x = \frac{1}{2}$).Since $x \in [0, 1]$,the maximum value occurs at the boundaries $x=0$ or $x=1$,which is $1$.Therefore,$\frac{3}{4} \le A \le 1$.

If $A = \sin^2 \theta + \cos^4 \theta$,then for all real values of $\theta$:

Similar Questions

$\frac{{\sin (B + A) + \cos (B - A)}}{{\sin (B - A) + \cos (B + A)}} = $

In the graph of the function $\sqrt{3} \sin x + \cos x$,the maximum distance of a point from the $x$-axis is:

If $y = (1 + \tan A)(1 - \tan B)$ where $A - B = \frac{\pi}{4}$,then $(y + 1)^{y + 1}$ is equal to

$\frac{\cos A}{1 - \sin A} = $

In any triangle $ABC$,the expression ${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module