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

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

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.

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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 $\tan\theta=\sqrt{\frac35}$.

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