Show that: $f(\theta)=\sin\theta\cos(\theta\ -k)$ is max when $\theta = \frac{k+90^{\circ}}{2}$ without using calculus.

Question

Given $$f(\theta)=\sin\theta\cos(\theta\ -k)$$

Show that $f(\theta)$ is maximum when: $\theta = \frac{k+90^{\circ}}{2}$

I can do this easily using calculus, but I'm looking for a way of doing it without calculus.

Context:

A particle is projected up an inclined slope. The incline is fixed at an angle $k$ to the horizontal. The particle is projected at an angle $\theta$ to the incline. This problem resulted from trying to find the angle of maximum range.

Vishu · Accepted Answer · 2020-07-12T17:16:02.873

3

Note $$\sin\theta \cos(\theta-k) = \frac 12\cdot 2\sin\theta\cos(\theta-k) =\frac 12 \left(\sin(2\theta-k) +\sin k \right) $$ This will be maximum, when $$\sin(2\theta-k) =1 \\ \implies 2\theta-k=\frac{\pi}{2} +2n\pi \\ \implies \theta=\frac 12\left (k+\frac{\pi}{2}+2n\pi \right)$$ for any integer $n$. $|\theta|$ can be minimized by setting $n=0$.

edited Jul 12 '20 at 17:16

answered Jul 12 '20 at 17:10

Vishu

14,469

2

+1 Why has it been downvoted? – Koro Jul 12 '20 at 17:14
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-@Koro, correct. It happened with me, too. It might be because OP didn't provide context and still @Tavish has answered it. – SarGe Jul 12 '20 at 17:31
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@SarGe Yet the post is not closed and has been upvoted $3$ times? – Vishu Jul 12 '20 at 17:39

Show that: $f(\theta)=\sin\theta\cos(\theta\ -k)$ is max when $\theta = \frac{k+90^{\circ}}{2}$ without using calculus.

1 Answers1