Showing $\cos(P − Q) = \frac{33}{85}$ for obtuse $P$ and reflex $Q$ such that $\sin P = \frac{8}{17}$ and $\tan Q = \frac43$,

Question

$\sin P = \frac{8}{17}$ and $\tan Q = \frac43$. If $P$ is obtuse and $Q$ is reflex, show clearly that $\cos(P − Q) = \frac{33}{85}$.

My working:

$$\cos P \times \cos Q + \sin P \times \sin Q$$ $$\left(-\frac{15}{17} \times -\frac35\right) + \left(\frac{8}{17} \times -\frac45\right) = \frac{13}{85}$$ which is not $33/85$. Where have I gone wrong?

Answer 1

$\cos(p-q) =\cos(p)×\cos(q)+\sin(p)×\sin(q)$

$\Rightarrow$$\cos(p-q)= \cos(q) (\cos(p) +\sin(p) ×\tan(q)) $

$\Rightarrow$$\cos(p-q) =\cos(q) (\cos(p) +\frac{8}{17}×\frac{4}{3})$

We know $\sin(p) =\frac{8}{17}$

$\Rightarrow$$1-\sin(p)^2=1-(\frac{8}{17})^2$

$\Rightarrow$$ \cos(p) =\frac{15}{17}$ or $\cos(p) =-\frac{15}{17}$

And we know $\tan(q) =\frac{4}{3}$$\Rightarrow$$ 1+\tan(q)^2=1+(\frac{4}{3})^2$$\Rightarrow$$ \cos(q) =\frac{3}{5} $ or $ \cos(q) =-\frac{3}{5}$

$\cos(p-q) =\cos(q) (\cos(p) +\frac{8}{17}×\frac{4}{3})>0$$\Rightarrow $$ \cos(q) >0 $ and $\cos(p) >\frac{-32}{51}$ So In this Cas $\cos(q) =\frac{3}{5}$ and $ \cos (p) =\frac{15}{17}$

So $\cos(p-q) =\frac{3}{5} (\frac{15}{17}+\frac{8}{17}×\frac{4}{3})=\frac{77}{85}$

In other Case $\cos(q) <0 $ and $\cos(p) <\frac{-32}{51}$ so $\cos(q) =-\frac{3}{5}$ and $\cos(p) =\frac{-15}{17}$

So $\cos(p-q) =\frac{-3}{5} (\frac{-15}{17}+\frac{8}{17}×\frac{4}{3})=\frac{13}{85}$