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

Answer: (D)

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

Answer 2

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

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

Since, $–1 < \sin(\theta + \varphi) < 1,$

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