Sin and Cos without a calculator

eli140

How do I use the unit circle to find sin and cos of an angle?
Thanks,

**Posted:** Sun Nov 08, 2009 9:20 am

AndrewTom

If the line segment (radius) joining the point $P(x, y)$ on the circle to the origin (the centre of the circle) makes an angle of $\theta$ with the positive direction of the $x-$ axis, then $x= \cos \theta$ and $y= \sin \theta$ , by Pythagoras.

**Posted:** Sun Nov 08, 2009 9:34 am

tornado.adv4 · Poincare Conjecture Offline Joined: 06 May 2009 Posts: 107

It's useful for finding values like $\sin0^{\circ}$ , $\sin90^\circ$ , $\sin180^\circ$ , and $\sin270^\circ$ (and cos, tan, etc. of those angles too).

For example, for $\sin90^\circ$ , you would look at a $90^\circ$ angle and see that it goes through the point (0,1) on the unit circle. The y-coordinate is 1, so $\sin90^\circ=1$ .

For values like $\sin30^\circ$ , $\sin45^\circ$ , and $\sin60^\circ$ , you would draw a right triangle in the unit circle that goes from (0,0) to (x,y) to (x,0).
The hypotenuse is the radius of the unit circle, so it is 1.

For finding $\sin30^\circ$ or $\sin60^\circ$ , you would use the relationships from a 30-60-90 triangle to find the side lengths of the right triangle you drew.
With $30^\circ$ , you would have a triangle with hypotenuse 1, adjacent side $\sqrt{3}/2$ , and opposite side 1/2. Then $\sin30^\circ=1/2$ , and $\cos30^\circ=\sqrt{3}/2$ .
For finding $\sin45^\circ$ , you would use the relationships from a 45-45-90 triangle.

**Posted:** Sun Nov 08, 2009 9:51 am

eli140

So to find sin and cos of an odd angle. say like 37 degrees, i would just have to graph it?

**Posted:** Sun Nov 08, 2009 10:11 am

pascal12 · Riemann Hypothesis Offline Joined: 18 Mar 2009 Posts: 253

Actually, 37 and 53 degrees are two more special ones. Although it is not exact, sin37 is approximately 3/5 and cos 37 is approximately 4/5.

You can't graph a triangle by hand and find the sinusoidal functions accurately.

**Posted:** Sun Nov 08, 2009 10:20 am

ernie

You can also use the Sum-Difference formulas, which say that $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha$ , and that $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ . The link provided shows a proof.

**Posted:** Sun Nov 08, 2009 10:33 am

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