If $\cos\alpha=\frac13,$ find the value of $7\cos(180^\circ-\alpha)-2\sin(90^\circ+\alpha)$ If $\cos\alpha=\dfrac13,$ find the value of $$7\cos(180^\circ-\alpha)-2\sin(90^\circ+\alpha)$$
Let $A=7\cos(180^\circ-\alpha)-2\sin(90^\circ+\alpha)=-7\cos\alpha-2\cos\alpha=-9\cos\alpha$ or when $\cos\alpha=\dfrac13\Rightarrow A=-9.\dfrac13=-3.$ The given answer in my book is $3$. Am I wrong?

$7 \cos(\pi-\alpha) = -7 \cos(\alpha) = -7/3$
$2 \sin(\pi/2 + \alpha) = 2 \cos (\alpha) = 2/3$
So you get $-7/3-2/3 = -3$.
