The value of a cos θ + b sin θ lies between

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Question

The value of a cos $θ$ + b sin $θ$ lies between

Options:

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: Option D
:
D

a cos $θ$ + b sin $θ$ = $\sqrt{a^{2} + b^{2}}$ $(\frac{a c o s θ}{\sqrt{a^{2} + b^{2}}} + \frac{b s i n θ}{\sqrt{a^{2} + b^{2}}})$

= $\sqrt{a^{2} + b^{2}}$ $s i n (θ + ϕ)$

Since, $- 1 \leq s i n (θ + ϕ)$ $\leq$ 1,

Then - $\sqrt{a^{2} + b^{2}}$ $\leq$ $s i n (θ + ϕ)$ $\leq$ $\sqrt{a^{2} + b^{2}}$ .

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