# Given tan θ and cos θ (below), solved for sinθ but got multiple answers

$$\tan \theta = \frac{\sqrt15}{3}$$

$$\cos \theta=\frac{\sqrt10}{4}$$

We're supposed to solve for $$\sin \theta$$ using trig identities, so:

### Pythagorean

$$\sin^2{\theta}+\cos^2{\theta}=1$$

$$\sin^2{\theta}+\left(\frac{\sqrt10}{4}\right)^{2}=1$$

$$\sin^2{\theta}+\frac{10}{16}=1$$

$$\sin^2{\theta}=1-\frac{10}{16}$$

$$\sin^2{\theta}=\frac{6}{16}=\frac{3}{8}$$

$$\sin{\theta}=\sqrt{\frac{3}{8}}=\frac{\sqrt{24}}{8}=\frac{\sqrt{6}}{4}$$

### sin/cos=tan

$$\frac{\sin\theta}{\cos\theta}=\tan\theta$$

$$\frac{\sin\theta}{\frac{\sqrt10}{4}}=\frac{\sqrt15}{3}$$

$$\sin{\theta}=\frac{\sqrt10}{4}*\frac{\sqrt15}{3}$$

$$\sin{\theta}=\frac{\sqrt150}{12}=\frac{5\sqrt6}{12}$$

Using this method, $$\sin{\theta}$$ is somehow greater than one? The above answer simplifies to about $$1.0206$$.

### 1 + cot$$^2\theta$$ = csc$$^2\theta$$

$$1+\cot^2{\theta}=\csc^2{\theta}$$

$$1+\left(\frac{3}{\sqrt15}\right)^2=\csc^2{\theta}$$

$$1+\frac{9}{15}=\csc^2{\theta}$$

$$\frac{24}{15}$$ = csc$$^2\theta$$

$$\frac{15}{24}=\sin^2{\theta}$$

$$\sqrt{\frac{15}{24}}=\sin{\theta}$$

$$\sqrt{\frac{5}{8}}=\sin{\theta}$$

$$\frac{\sqrt40}{8}=\sin{\theta}$$

$$\frac{2\sqrt10}{8}$$ = sin $$\theta$$

$$\frac{\sqrt10}{4}=\sin{\theta}$$

This is the $$\cos$$ value (maybe I made a mistake?)

I have asked my math teacher about this. She is also very confused as to what the answer is: the worksheet she printed out wasn't hers, and the answer listed is the second one, which is greater than one and is probably wrong somehow. I realized I could use the third identity only after I met with her, but she couldn't find any problem with the first two methods, and yet somehow they yielded different values.

Could someone please shed some light as to why I ended up with multiple answers, with one greater than one? Thanks for any answers.

The problem has no solutions because there is no $$\theta$$ such that $$\tan\theta=\frac{\sqrt{15}}3$$ and that $$\cos\theta=\frac{\sqrt{10}}4$$. In fact, if $$\cos\theta=\frac{\sqrt{10}}4$$, then$$\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=\frac{1-\cos^2\theta}{\cos^2\theta}=\frac{\frac6{16}}{\frac{10}{16}}=\frac35,$$and therefore $$\tan\theta=\sqrt{\frac35}$$.