# Thread: solve 8 sin (theta)-15cos(theta)=8.5

1. ## solve 8 sin (theta)-15cos(theta)=8.5

i am trying to workout part (ii) illustrate your method

i already have done part (i)

(i) Express $\displaystyle 8 Sin\theta - 15Cos\theta$

in the format $\displaystyle R.Sin(\theta + \alpha)$

$\displaystyle \theta = tan^{-1} \frac {15}{8}$

$\displaystyle \theta = 61.9^0$

$\displaystyle \theta$=360-61.9 = $\displaystyle 298.1^0$

$\displaystyle 8 Sin\theta - 15Cos\theta = 8.5$

for $\displaystyle 0^0\leq\theta\leq360^0$

any help much appreciated

2. Hello, decoy808!

i already have done part (i) . . . . but you didn't finish

(i) Express: .$\displaystyle X \:=\:8\sin\theta - 15\cos\theta$ in the format: $\displaystyle R\sin(\theta + \alpha)$
This is how I would do it . . .

We have: .$\displaystyle \sqrt{8^2+15^2} \:=\:\sqrt{289} \:=\:17$

Multiply $\displaystyle X$ by $\displaystyle \tfrac{17}{17}\!:\;\;\frac{17}{17}\left(8\sin\thet a - 15\cos\theta\right) \;=\;17\left(\tfrac{8}{17}\sin\theta - \tfrac{15}{17}\cos\theta\right)$

Let $\displaystyle \alpha$ be in a right triangle with: .$\displaystyle opp = 15,\:adj= 8,\:hyp = 17$
. . Then: .$\displaystyle \cos\alpha = \tfrac{8}{15},\;\sin\alpha = \tfrac{15}{17}$

Then we have: .$\displaystyle X \;=\;17(\cos\alpha\sin\theta - \sin\alpha\cos\theta)$

From a Compound-Angle Identity: .$\displaystyle \boxed{X \;=\;17\sin(\theta - \alpha)}$

Since $\displaystyle \sin\alpha \:=\: \tfrac{15}{17}\:=\:0.882352941$, then: .$\displaystyle \alpha \,\approx\,61.9^i$

(ii) Using your answer to part (i), solve: .$\displaystyle 8\sin\theta - 15\cos\theta \:=\: 8.5\;\;\text{ for }0^o\leq\theta\leq360^o$

So we have: .$\displaystyle 8\sin\theta -15\cos\theta \:=\:8.5$

. . which becomes: .$\displaystyle 17\sin(\theta - 69.1^o) \:=\:8.5$

Then: .$\displaystyle \sin(\theta - 61.9^o) \:=\:0.5\quad\Rightarrow\quad \theta - 61.9^o \:=\:30^o,\:150^o$

. . Therefore: .$\displaystyle \boxed{\theta \;=\;91.9^o,\;211.9^o}$