Solving Trig Function with both $\sin$ and $\cos$

Question

I am trying to understand the function $h(t) = 65 + 36\sin\left(\frac{3t}{2}\right) - 15\cos\left(\frac{3t}{2}\right)$, where $t$ is the time in seconds and $\frac{3t}{2}$ is expressed in radians.

A question asks that I find the times in the first revolution when the height is exactly $65$m. Using basic algebra I have been able to solve in order to get time which is $t = 0.26319$. But the question asks for times in the first revolution and I am having trouble understanding how to find these values.

Hence, I need to find the time to complete one revolution and the angular speed of the function.

How would you determine the times when the height is $65$m and from that be able to determine the time it takes to complete one revolution and the angular speed of the function?

I assume that this equation is not in the form of $a + b\sin(cx)$?

Answer 1

You are looking for $t$ such that

$$ \begin{align} &\quad h(t) = 65\\ \Rightarrow &\quad 36\sin\left(\frac{3t}{2}\right) = 15\cos\left(\frac{3t}{2}\right)\\ \Rightarrow &\quad 12\sin\left(\frac{3t}{2}\right) = 5\cos\left(\frac{3t}{2}\right)\\ \Rightarrow &\quad 12^2\sin^2\left(\frac{3t}{2}\right) = 5^2\cos^2\left(\frac{3t}{2}\right)\\ \Rightarrow &\quad 144\sin^2\left(\frac{3t}{2}\right) = 25\left({1-\sin^2\left(\frac{3t}{2}\right)}\right)\\ \Rightarrow &\quad 169\sin^2\left(\frac{3t}{2}\right) = 25\\ \Rightarrow &\quad 13\sin\left(\frac{3t}{2}\right) = 5\\ \Rightarrow &\quad \sin\left(\frac{3t}{2}\right) = \frac{5}{13}\\ \Rightarrow &\quad \frac{3t}{2} = \sin^{-1}\left(\frac{5}{13}\right)\\ \Rightarrow &\quad t = \frac23\sin^{-1}\left(\frac{5}{13}\right) \end{align} $$

Answer 2

Hint

use $${a\over \sqrt{a^2+b^2}}\cos u+{b\over \sqrt{a^2+b^2}}\sin u=\cos(u-\theta)$$where$$\theta =\cos^{-1}{a\over \sqrt{a^2+b^2}}=\sin^{-1}{b\over \sqrt{a^2+b^2}}$$

Answer 3

The equation has the classical form

$$a\cos t+b\sin t=c.$$

Among million other ways, you can solve with

$$(a+b\tan t)^2=\frac{c^2}{\cos^2t}=c^2\tan^2t+c^2$$

or

$$(c^2-b^2)\tan^2t-2ab\tan t+c^2-a^2=0,$$

giving

$$\tan t=\frac{ab\pm c\sqrt{a^2+b^2-c^2}}{c^2-b^2}.$$