Let a be the maximum value of (3 cos theta - | vedclass

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

Question

(C) The maximum value of $3 \cos 	heta - 4 \sin 	heta$ is given by $\sqrt{3^2 + (-4)^2} = 5$. So,$a = 5$. Given $\alpha = 5 \sin^2 	heta \cos^3 	h

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The maximum value of $3 \cos 	heta - 4 \sin 	heta$ is given by $\sqrt{3^2 + (-4)^2} = 5$. So,$a = 5$. Given $\alpha = 5 \sin^2 	heta \cos^3 	heta$ and $\beta = 5 \sin^3 	heta \cos^2 	heta$. Then $\alpha^2 + \beta^2 = 25 \sin^4 	heta \cos^6 	heta + 25 \sin^6 	heta \cos^4 	heta = 25 \sin^4 	heta \cos^4 	heta (\cos^2 	heta + \sin^2 	heta) = 25 \sin^4 	heta \cos^4 	heta$. Therefore,$(\alpha^2 + \beta^2)^5 = (25 \sin^4 	heta \cos^4 	heta)^5 = (5^2 \sin^4 	heta \cos^4 	heta)^5 = 5^{10} \sin^{20} 	heta \cos^{20} 	heta$. Also,$\alpha \beta = (5 \sin^2 	heta \cos^3 	heta)(5 \sin^3 	heta \cos^2 	heta) = 25 \sin^5 	heta \cos^5 	heta$. Then $(\alpha \beta)^4 = (25 \sin^5 	heta \cos^5 	heta)^4 = (5^2 \sin^5 	heta \cos^5 	heta)^4 = 5^8 \sin^{20} 	heta \cos^{20} 	heta$. Finally,$\sqrt{\frac{(\alpha^2 + \beta^2)^5}{(\alpha \beta)^4}} = \sqrt{\frac{5^{10} \sin^{20} 	heta \cos^{20} 	heta}{5^8 \sin^{20} 	heta \cos^{20} 	heta}} = \sqrt{5^2} = 5$.

Let $a$ be the maximum value of $(3 \cos \theta - 4 \sin \theta)$ and $\theta \neq \frac{n \pi}{2}$. If $\alpha = a \sin^2 \theta \cos^3 \theta$ and $\beta = a \sin^3 \theta \cos^2 \theta$,then $\sqrt{\frac{(\alpha^2 + \beta^2)^5}{(\alpha \beta)^4}} = $

Similar Questions

Let $f(x) = \cos 5x + A \cos 4x + B \cos 3x + C \cos 2x + D \cos x + E$,and $T = f(0) - f\left(\frac{\pi}{5}\right) + f\left(\frac{2\pi}{5}\right) - f\left(\frac{3\pi}{5}\right) + \dots + f\left(\frac{8\pi}{5}\right) - f\left(\frac{9\pi}{5}\right)$. Then,$T$

If $A, B, C$ are the angles of a triangle,then $\sin 2A - \sin 2B + \sin 2C =$

If $A+B=C$,then the value of $\cos ^2 A+\cos ^2 B+\cos ^2 C-2 \cos A \cos B \cos C$ is equal to:

The extreme values of $1 + 4 \sin \theta + 3 \cos \theta$ are

The maximum value of the function $f(x) = 3 \sin^{12} x + 4 \cos^{16} x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module