I tried to solve this trig identity problem from my Trigonometry textbook and I can't seem to get the correct answer. I've solved similar trig identity problems and have successfully found the correct answers for other problems; however, I'm not familiar with being given "$\sin(\alpha/2)$". (I have the answers for this math question, but my textbook doesn't give an explanation for how to solve this particular problem).
Using the half-angle or double-angle formula was recommended to me on another math forum, but I'm unsure how to properly apply those trigonometric formulas to this specific math problem. I'm used to applying the Pythagorean Identity "$\sin^2 x + \cos^2x = 1$" to tackle this kind of problem, but since I've never been given "$\sin(\alpha/2)$" for my homework assignments or quizzes I don't have proper notes to reference for this specific math problem. :/
Correct Answers from the textbook:
$$\sin a = -24/25$$
$$\cos a = 7/25$$
$$\tan a = -24/7$$
$$\csc a = -25/24$$
$$\sec a = 25/7$$
$$\cot a = -7/24$$
Any help is appreciated ~ Thank you 🙂
Best Answer
Let's begin by finding the value of $\cos(\frac{\alpha}{2})$.
$\sin(\frac{\alpha}{2}) = \frac{3}{5}$ so $\sin^2(\frac{\alpha}{2}) = \frac{9}{25}$.
$\cos^2(\frac{\alpha}{2}) + \sin^2(\frac{\alpha}{2}) = 1 \implies \cos^2(\frac{\alpha}{2}) = 1 - \frac{9}{25} = \frac{16}{25}$
$\frac{\alpha}{2} \in (\frac{3\pi}{4}, \pi)$ so $\cos(\frac{\alpha}{2}) < 0$.
As such, $\cos(\frac{\alpha}{2}) = - \sqrt{\frac{16}{25}} = -\frac{4}{5}$.
Now that we have both the values of $\sin$ and $\cos$. We can just use the half angle identities:
$\sin(\alpha) = 2 \sin(\frac{\alpha}{2}) \cdot \cos(\frac{\alpha}{2})$
And $\cos(\alpha) = \cos^2(\frac{\alpha}{2}) - \sin^2(\frac{\alpha}{2}) = 1 - 2 \sin^2(\frac{\alpha}{2})$. And deduce the rest from that.