# Use identities to find the exact values of the remaining five trigonometric functions at alpha. $\sin(\alpha/2) = 3/5$ and $3\pi/4 < \alpha/2 < \pi$

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 :)

• It might be helpful to let $\theta = \alpha/2$. Then the question is about finding the values of the functions at $2\theta$. See also the Wikipedia article on double-angle identities. Nov 16, 2020 at 18:04

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.

• Hi Oussema, thank you for the suggestion :) I understand the first step in using the Pythagorean Identity, but I'm lost on the half-angle identity step. How did you get sin(a)=2sin(a/2)*cos(a/2), and cos(a)=cos^2(a/2)-sin^2(a/2)=1-2sin^2(a/2)?
– user850185
Nov 18, 2020 at 23:42
• I'm applying the sin and cos of sum formulas. The most elementary proof of these formulas I know is geometric, and it's quite a bit long. There are shorter proofs with complex numbers and Euler's identity but I assume you do not know complex numbers. If you're interested in the geometric proof, take a look at this web page: themathpage.com/aTrig/sum-proof.htm Nov 20, 2020 at 13:25