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 $1$.
 A: Hints:


*

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

*Write $315$ as $270 + 45$ 
A: 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 $\tan{\theta} = \frac{\sin{\theta}}{\cos \theta}$.
A: Two more possibilities:


*

*Use a table of trigonometric values.

*Use a ruler and a protractor in drawing the line values of the trigonometric functions.  (See, for example, this.)

A: 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 
$\sin(-315)=-\sin(270+45)=\cos(45) = \frac{1}{\sqrt{2}}$
$\cos(-315)=\cos(315)=\cos(270+45)=\sin(45)  = \frac{1}{\sqrt{2}}$
$\tan(-315)=-\tan(270+45)=\cot(45) = 1$
Check this link for more info on converting trigo identities
A: 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 $45$
