Question

Let P be a point inside a triangle ABC such that PA=PC=AB and angles ABC, PAC and PCB are of magnitude 12$\alpha$, 3$\alpha$ and 2$\alpha$ respectively. Then maximum value of expression f($\theta$) = sin 5$\alpha$ sin$\theta$ + cos 10$\alpha$ cos$\theta$,($\theta$ ∈ R) is equal to,

• A. $\sqrt{2}$
• B. 1
• C. 2
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (D)

$\frac{1}{\sqrt{2}}$

Answer 2

1. Notice that the expression

$$f(\theta)=\sin5\alpha\sin\theta+\cos10\alpha\cos\theta,$$

can be written in the form

$$A\sin\theta+B\cos\theta,$$

whose maximum value (over $\theta\in\Bbb R$) equals

$$R=\sqrt{A^2+B^2}=\sqrt{\sin^2 5\alpha+\cos^2 10\alpha}\,.$$

2. In the given geometric configuration the point $P$ inside $\triangle ABC$ satisfies

$$PA=PC=AB.$$

By using the angles provided:

• $\angle ABC=12\alpha$,
• $\angle PAC=3\alpha$,
• $\angle PCB=2\alpha$, one may show (by splitting $\angle A$ of $\triangle ABC$ and also considering $\triangle ABP$ and $\triangle BPC$) that consistency forces

$$\alpha=6^\circ.$$

3. Then

$$5\alpha=30^\circ,\qquad \sin5\alpha=\sin30^\circ=\tfrac12,$$

and

$$10\alpha=60^\circ,\qquad \cos10\alpha=\cos60^\circ=\tfrac12.$$

4. Therefore,

$$R=\sqrt{\left(\tfrac12\right)^2+\left(\tfrac12\right)^2}=\sqrt{\tfrac{1}{4}+\tfrac{1}{4}}=\sqrt{\tfrac{1}{2}}=\tfrac1{\sqrt2}\,.$$

Thus, the maximum value of $f(\theta)$ is $\tfrac{1}{\sqrt2}$.