# Evaluate the $\sin$, $\cos$ and $\tan$ without using calculator?

Evaluate the $\sin$, $\cos$ and $\tan$ without using calculator?

$150$ degree

the right answer are $\frac{1}{2}$, $-\frac{\sqrt{3}}{2}$and $-\frac{1}{\sqrt{3}}$

$-315$ degree

the right answer are $\frac{1}{\sqrt{2}}$, $\frac{1}{\sqrt{2}}$ and $1$.

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You may find this helpful... – draks ... Apr 23 '12 at 8:50

You can look up cos and sin on the unit circle.

The angles labelled above are those of the special right triangles 30-60-90 and 45-45-90. Note that -315 ≡ 45 (mod 360).

For tan, use the identity $\tan{\theta} = \frac{\sin{\theta}}{\cos \theta}$.

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Hints:

• Write $150$ as $90 + 60$ and use $\sin(A+B),\cos(A+B),\tan(A+B)$ formulas.

• Write $315$ as $270 + 45$

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It is very simple.

$\sin(150) = \sin(90+60) = \cos(60) = \frac{1}{2}$
$\cos(150) = \cos(90+60) = - \sin(60) = -\frac{\sqrt{3}}{2}$ $\tan(150) = \tan(90+60) = - \cot(60) = -\frac{1}{\sqrt{3}}$

similarly

$\sin(-315)=-\sin(270+45)=\cos(45) = \frac{1}{\sqrt{2}}$
$\cos(-315)=\cos(315)=\cos(270+45)=\sin(45) = \frac{1}{\sqrt{2}}$
$\tan(-315)=-\tan(270+45)=\cot(45) = 1$

I would write $150=180-30$ and use $\cos (180-\theta)=-\cos \theta$ and so on. Then $-315=45-360$ so it is just the functions at $45$