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This is part of a physics problem I was doing yesterday. I am supposed to find the maximum value of $$\mu^{-1}\cos\theta+\sin\theta$$ This is is supposed to produce the result of $\sqrt{1+\mu^{-2}}$. However I am not getting that. By taking the derivative of the equation, and setting it equal to zero I got $$-\mu^{-1}\sin\theta+\cos\theta=0$$ $$\mu\cos\theta=\sin\theta$$ $$\mu=\tan\theta$$ I don’t get the same result of $\sqrt{1+\mu^{-2}}$ after plugging $\tan^{-1}{(\mu)}$ back into our equation. Can anyone help?

Physics_Learner
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2 Answers2

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It is known that for $A\cos(\theta) + B\sin(\theta)$, we can represent it as 1 trigonometric function: $\sqrt{A^2+B^2}\cos(\theta + \phi)$ or $\sqrt{A^2+B^2}\sin(\theta+\omega)$. Clearly it can be $\sin$ or $\cos$ because they are just shifted waves of each other.

From these expressions of 1 sinusoid, it is clear the maximum value is $\sqrt{A^2+B^2}$, giving your answer of $\sqrt{1+\mu^{-2}}$.

Using the calculus approach, we get that a critical point (other than 0) is $\arctan(\mu)$.

Plugging this in the original function we get $\frac{1}{\mu}\cos(\arctan(\mu))+\sin(\arctan(\mu))$.

Now it will help to draw a right triangle with legs $\mu$, $1$ and hypotenuse $\sqrt{1+\mu^2}$ because then there is an angle of the triangle equal to $\arctan(\mu)$.

Using this triangle, $\sin(\arctan(\mu))=\mu/\sqrt{1+\mu^2}$ and $\cos(\arctan(\mu))=1/\sqrt{1+\mu^2}$.

$$\mu^{-1}/\sqrt{1+\mu^2} + \mu/\sqrt{1+\mu^2} = \mu^{-1}(1+\mu^2)/\sqrt{1+\mu^2} = \mu^{-1}\sqrt{1+\mu^2}=\sqrt{1+\mu^{-2}},$$the desired result as the maximum value.

rogerl
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Abhishek Vangipuram
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You just need to simplify the result.

When you set $\theta = \tan^{-1}\mu$ you get $$\cos\theta = \frac{1}{\sqrt{1+\mu^2}},\quad \sin\theta = \frac{\mu}{\sqrt{1+\mu^2}},$$ so that $$\mu^{-1}\cos\theta + \sin\theta = \frac{\mu^{-1}+\mu}{\sqrt{1+\mu^2}} = \frac{1+\mu^2}{\mu\sqrt{1+\mu^2}} = \frac{\sqrt{1+\mu^2}}{\mu} = \sqrt{\mu^{-2}+1}.$$

rogerl
  • 22,399