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Say $\tan a = -7/24$ (second quadrant) and $\cot b = 3/4$ (third quadrant), how would I find $\sin (a + b)$?

I figured I could solve for the $\sin/\cos$ of $a$ & $b$, and use the add/sub identities, but I got massive unwieldy numbers outside the range of $\sin$. How ought I to go about this problem? Thanks.

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  • $\begingroup$ Use all sin tan cos rule to determine the signs $\endgroup$ – lab bhattacharjee Apr 30 '16 at 17:52
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$\tan(a) = -7/24$

Opposite side $= 7$ and adjacent side $= 24$

Pythagorean theorem

$\Rightarrow$ hypotenuse $= \sqrt{49+576} = 25$

$\sin(a) = 7/25$ (sin is positive in second quadrant)

$\cos(a) = - 24/25$ (cos is negative in second quadrant)

$\cot(b) = 3/4 \Rightarrow \tan(b) = 4/3$

Opposite side $= 4$ and adjacent side $= 3$

Pythagorean theorem

$\Rightarrow$ hypotenuse $= \sqrt{9+16} = 5$

$\sin(b) = - 4/5$ (sin is negative in third quadrant)

$\cos(b) = -3/5$ (cos is negative in third quadrant)


$\sin(a+b) = \sin a \cos b + \cos a \sin b$

$\sin(a+b) = 3/5$

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HINT:

$$-\dfrac7{24}=\tan a\iff\dfrac{\sin a}7=\dfrac{\cos a}{-24}=\pm\sqrt{\dfrac1{7^2+24^2}}=\pm\dfrac1{25}$$

As $a$ is in the second qaudrant, $\sin a>0,\cos a<0$

Can you manage $\cot b=\dfrac34?$

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