Question

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

• A. a-b and a+b
• B. a and b
• C. -$(a^2+b^2)$ And $(a^2+b^2)$
• D. -$\sqrt{a^2+b^2}$ And $\sqrt{a^2+b^2}$

Answer 1

The correct option is D

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

a cos $\theta$ + b sin $\theta$ = $\sqrt{a^2+b^2}$ $(\frac{acos\theta }{\sqrt{a^2+b^2}}+\frac{bsin\theta }{\sqrt{a^2+b^2}})$

= $\sqrt{a^2+b^2}$ $sin(\theta +\varphi )$

Since, $-1\le sin(\theta +\varphi )$ $\le$ 1,

Then -$\sqrt{a^2+b^2}$ $\le$ $sin(\theta +\varphi )$ $\le$ $\sqrt{a^2+b^2}$.