# 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 $8 Sin\theta - 15Cos\theta$

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

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

$\theta = 61.9^0$

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

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

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

any help much appreciated

2. Hello, decoy808!

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

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

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

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

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

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

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

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

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

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

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

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

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