The value of a cos theta + b sin theta lies b | vedclass

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

Question

(D) Let $f(	heta) = a \cos 	heta + b \sin 	heta$.We can write this as $f(	heta) = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \cos 	heta +

Vedclass · Accepted Answer

$-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$

Vedclass · Answer

(D) Let $f(	heta) = a \cos 	heta + b \sin 	heta$.We can write this as $f(	heta) = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \cos 	heta + \frac{b}{\sqrt{a^2 + b^2}} \sin 	heta ight)$.Let $\cos \phi = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin \phi = \frac{b}{\sqrt{a^2 + b^2}}$.Then $f(	heta) = \sqrt{a^2 + b^2} (\cos 	heta \cos \phi + \sin 	heta \sin \phi) = \sqrt{a^2 + b^2} \cos(	heta - \phi)$.Since the range of $\cos(	heta - \phi)$ is $[-1, 1]$,the range of $f(	heta)$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Therefore,the value of $a \cos 	heta + b \sin 	heta$ lies between $-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$.

The value of $a \cos \theta + b \sin \theta$ lies between

Similar Questions

The range of $f(x) = \cos x - \sin x$ is

The maximum value of $12 \sin \theta - 9 \sin^2 \theta$ is

If in a $\Delta ABC$,$\sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \cdot \sin B \cdot \sin C$,then:

Maximum value of $f(x) = \sin x + \cos x$ is

If $\sin \theta = \frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} \right)$,where $x, y \in \mathbb{R} - \{0\}$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module