If $3 \sin \theta + 4 \cos \theta = 5$, find the value of $\sin \theta$.

Question

What would be the answer for the following question?

If $3 \sin \theta + 4 \cos \theta = 5$, find the value of $\sin \theta$.

Answer 1

Hint:

Rewrite the equation as $\quad 4\cos\theta=5-3\sin \theta$.

Now $ \cos\theta=\pm\sqrt{1-\sin^2\theta} $, so setting $\:s=\sin\theta$, we obtain the irrational equation $$\pm4\sqrt{1-s^2}=5-3s \iff 16(1-s^2)=(5-3s)^2.$$ Of course, one possibly has to eliminate the solution(s) of this equation such that $\;|s|>1$.

Answer 2

One approach you can use, that keeps things relatively simple, is to do this:

First, rearrange the equation to $4 \cos \theta = 5 - 3 \sin \theta$.

Then, square both sides. On the left, you'll have $16 \cos^2 \theta$, which you can replace with $16(1 - \sin^2 \theta)$, and you can then reorganise everything into a quadratic equation in terms of $\sin \theta$.

Once you've solved that, though, you'll need to check the two answers you get, because one of them will not be a solution to the original equation (it gets added to the system when you square everything).

Answer 3

Hint:

Consider a right triangle $ABC$, with right angle in $A$, and with $a=5, \; b=3, \; c=4$.

Then $\theta$ is the angle ...

Answer 4

There are various ways to solve the general problem of $$a \sin \theta + b \cos \theta = c$$

Way 1

A cool way to solve it is using the tan-half-angle substitution $t = \tan \left( \frac{\theta}{2} \right)$ with $$ \begin{aligned} \sin \theta & = \tfrac{2 t}{1+t^2} & \cos \theta & = \tfrac{1-t^2}{1+t^2} \end{aligned} $$

This turns the trig problem into a polynomial

$$ a \tfrac{2 t}{1+t^2} + b \tfrac{1-t^2}{1+t^2} = c $$ $$ 2 t a + (1-t^2) b = (1+t^2) c $$ $$ t = \frac{a \pm \sqrt{a^2 +b^2-c^2}}{b+c} $$

and finally $$ \boxed{ \theta = 2 \tan^{-1}(t) }$$

Way 2

Split the angle into two parts $\theta = x +y $ and expand the sine and cosine

$$ \sin \theta = \cos x \sin y + \sin x \cos y $$ $$ \cos \theta = \cos x \cos y - \sin x \sin y $$

Together you have

$$ a \left(\cos x \sin y + \sin x \cos y\right) + b \left(\cos x \cos y - \sin x \sin y\right) = c $$

which you re-arrange as

$$ \cos x \underbrace{ \left( b \cos y + a \sin y \right)}_{\text{make zero} } + \sin x \left(a \cos y -b \sin y \right) = c $$

Solve as two equations with $$b \cos y + a \sin y = 0 $$

$$ y = -\tan^{-1} \left(\tfrac{b}{a}\right) $$

and

$$ \sin x \left( a \cos y + b \sin y \right) = c $$

$$ \sin x \sqrt{a^2+b^2} = c $$

$$ x = \sin^{-1} \left( \tfrac{c}{\sqrt{a^2+b^2}} \right) $$

Or altogether

$$ \boxed{ \theta = \sin^{-1} \left( \tfrac{c}{\sqrt{a^2+b^2}} \right) -\tan^{-1} \left(\tfrac{b}{a}\right) } $$