# Thread: Trigonometry Help

1. ## Trigonometry Help

I have done part of this question but need help finishing it off

also first time i am using the math code so excuse if i go wrong

$I={4\sin\theta}-{3\cos\theta}$
$I={R\sin(\theta-\alpha)}$

Part (a) was to find R and $\alpha$ for which i got
$R = 5$
$\alpha = 36.9$

Part (b) Hence solve the equation ${4\sin\theta}-{3\cos\theta}=3$ to find values of $\theta$ between 0 and 360

Part (c) write down greatest value for I

2. Hello, Stylis10!

$I\:=\:{4\sin\theta}-{3\cos\theta}$
$I\:=\:R\sin(\theta-\alpha)$

(a) Find $R$ and $\alpha$

I got: . $R \:= \:5,\;\;\alpha \:\approx \:36.9^o$ . . . . . Good!
(b) Hence solve the equation: . $4\sin\theta-3\cos\theta \:=\:3 \;\;\text{ for }0^o \leq \theta < 360^o$
Divide by 5: . $\frac{4}{5}\sin\theta - \frac{3}{5}\cos\theta \:=\:\frac{3}{5}$

Let $\tan\alpha \:=\:\frac{3}{4}\quad\hdots\quad\text{Then: }\:\cos\alpha \:=\:\frac{4}{5},\;\;\sin\alpha \:=\:\frac{3}{5}\:=\:0.6$

Then we have: . $\cos\alpha\sin\theta - \sin\alpha\cos\theta \:=\:\frac{3}{5}\quad\Rightarrow\quad \sin(\theta -\alpha) \:=\:0.6$

. . $\theta - \alpha \;=\;\sin^{-1}(0.6) \quad\Rightarrow\quad\theta \;=\;\sin^{-1}(0.6) + \alpha$

Since $\alpha \:=\:\sin^{-1}(0.6)$, we have: . $\theta \;=\;2\sin^{-1}(0.6)$

Since $\sin^{-1}(0,6) \:\approx\:36.9^o,\;143.1^o,\;\;\theta \;=\;73.8^o,\;286.2^o$

(c) Find the greatest value for $I.$
In part (a), we had: . $I \;=\;5\sin(\theta-36.9^o)$

The maximum value is: . $I \:=\:5$

3. thank you