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If $\cos(x)= -\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$ for $\frac{\pi}{2} < x < \pi$ and $\pi < y <\frac{3\pi}{2}$. What is the value of $\sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x) \cos(y) - \sin(x) \sin(y)$?


Solution:

If $\frac{\pi}{2} < x < \pi$ then right triangle with acute angle $x$ facing west at second quadrant, the hypotenuse must be positive and the only negative is the adjacent. $\cos(x)= -\frac{24}{25}$ then $adjacent=-24$ and $front=7$ so $\sin(x) = \frac{7}{25}$.

If $\pi < y <\frac{3\pi}{2}$ then right triangle with acute angle $x$ facing west at third quadrant, the front and adjacent are negative. If $\tan(y) = \frac{9}{40}$ then the front is $-9$, the adjacent is $-40$, and the hypotenuse is $41$. So $\sin(y) = -\frac{9}{41}$ and $\cos(y)=-\frac{40}{41}$

The right triangle that we consider is the one with the adjacent side on the $x$-axis right..?

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    $\begingroup$ Everything you are doing is correct. So all you have to do is plug in your values. $\endgroup$ – Zarrax Aug 22 '19 at 5:24
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Your method is correct. Indeed, refer to the graphs:

$\hspace{1cm}$enter image description here

$$\cos x=-\frac{24}{25}=\frac{\overbrace{-24}^{adjacent}}{25}, x\in \left(\frac{\pi}{2},\pi\right); \quad \tan y=\frac{9}{40}=\frac{\overbrace{-9}^{front}}{\underbrace{-40}_{adjacent}},y\in \left(\pi,\frac{3\pi}{2}\right).$$ Also note: $$\cos x=-\cos(\pi -x)=-\frac{24}{25} \Rightarrow \cos (\pi -x)=\frac{24}{25}; \quad \sin (\pi -x)=\frac{7}{25}=\sin x.\\ \tan y=\tan(y-\pi)=\frac{9}{40}=\frac{-9}{-40} \Rightarrow \begin{cases}\sin y=-\sin(y-\pi)=-\frac{9}{41} \Rightarrow \sin (y-\pi)=\frac9{41}\\ \cos y=-\cos(y-\pi)=-\frac{40}{41} \Rightarrow \cos (y-\pi)=\frac{40}{41}\end{cases}$$

Alternatively, you can remember the signs of sine, cosine, tangent and cotangent functions in the four quarters: sine (++--); cosine (+--+); tangent and cotangent (+-+-).

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You can observe that $$ \lvert\sin x\rvert=\sqrt{1-\cos^2x}=\sqrt{1-\frac{24^2}{25^2}}=\frac{7}{25} $$ Since $\pi/2<x<\pi$, we conclude that $\sin x=7/25$.

Also $$ \lvert\cos y\rvert=\sqrt{\frac{1}{1+\tan^2y}}=\frac{40}{41} $$ and from $\pi<y<3\pi/2$ we conclude that $\cos y=-40/41$. Thus $$ \sin y=\tan y\cos y=-9/41 $$

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