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I need help with solving this problem.

Find the value of $\sin2x$, if $\cot x = -\frac{7}{24}$, and $x$ is obtuse.

My attempt to this problem, was as follows :

First I knew that $\cot x$ equals a negative value, so the terminal arm must lie in the $2$-nd or $4$-th quadrant. I made the terminal arm be in the second quadrant, which will make $x$ be an obtuse angle. Then I found my hypotenuse, which is $25$, then I wrote the trigonometric ratios for cosine and sine. Which were : $\cos x = -\frac{7}{25}$ , $\sin x = \frac{24}{25}$. I then used the double angle formula for sine, to substitute my ratios in it, and solve. $$\sin 2x = 2\sin x \cos x$$, and I got $$\sin 2x = -2 \cdot \frac{24}{25} \cdot \frac{7}{25} = -\frac{366}{625}$$

Did I solve this problem correctly ?

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  • $\begingroup$ As $$\sin2x=\cdots=\dfrac{2\cot x}{1+\cot^2x}$$ What is the use of "obtuse"? $\endgroup$ – lab bhattacharjee Jan 10 '16 at 14:33
  • $\begingroup$ @labbhattacharjee This is what it said in the question. $\endgroup$ – Viktor Raspberry Jan 10 '16 at 14:40
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Yes, your answer is correct. It's nice that the problem made use of the $7, 24, 25$ Pythagorean triple.

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