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

The equation $(a + b)^2 = 4ab \sin^2 \theta$ is possible only when

$\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}+\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=$

If $0 < x < \pi$ and $\cos x + \sin x = \frac{1}{2}$,then $\tan x$ is:

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

If $\tan A = 2\tan B + \cot B,$ then $2\tan (A - B) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module