# Math Help - how to turn sin 45degrees to radical form?

1. ## how to turn sin 45degrees to radical form?

I have some homework and I do have the solutions manual that tells me sin45 degrees is 1/radical2. what I would like to know is how do I do this conversion with pen and paper and caculator?

2. Originally Posted by wonderd
I have some homework and I do have the solutions manual that tells me sin45 degrees is 1/radical2. what I would like to know is how do I do this conversion with pen and paper and caculator?
Consider a right isosceles triangle with two sides of length 1.

3. im not sure i understand, i would like to know how to do with with all angles

4. Originally Posted by wonderd
im not sure i understand, i would like to know how to do with with all angles
You asked how to get sin(45 degrees) and I have told you what to do. If you wanted other angles, you should have said so in your first post. I suggest you consult any standard textbook that covers this material, or use Google.

Regarding getting sin(45 degrees), did you draw the triangle? Do you know Pythagoras' Theorem? Have you studied trigonometry? If so, then what part of my suggestion are you having trouble with?

5. can someone please show me the exact step on how to do this for ANy angle? i have searched google

6. Originally Posted by wonderd
can someone please show me the exact step on how to do this for ANy angle? i have searched google

The value of sine and cosine of special angles such as 30, 45 and 60 degrees can be found using triangles (in a similar way to what I have already explained). For angles such as 15 degrees, the compound angle formulae can be used since 15 = 60 - 45.

There are a multitude of techniques for other angles.

7. For the great majority of angles, you cannot write sine or cosine in terms of radicals. There are a few for which you can.

For any right triangle, the two non-right angles must add to 90 degrees.

If one angle is $\alpha= 45$ then the other is $\beta= 90- 45= 45$ also. Since the two angles are the same, the two legs must have the same length (that is the "isosceles" triangle mr fantastic mentioned). Take the two legs to be of length "s". By the pythagorean theorem, the hypotenuse has length c given by $c^2= s^2+ s^2= 2s^2$ so $c= s\sqrt{2}$. Then "sine= opposite/hypotenuse", $sin(45)= \frac{s}{s\sqrt{2}}= \frac{1}{\sqrt{2}}$ and "cosine= near side/hypotenuse", so $cos(45)= \frac{s}{s\sqrt{2}}= \frac{1}{\sqrt{2}}$.

An "equilateral" triangle has all three sides the same length and all three angles the same. Since the angles in any triangle add to 180 degrees, each of the three equal triangles has measure 180/3= 60. Now draw a perpendicular from one vertex to the opposite side. By symmetry that divides the triangle into two identical right triangles. One of the angles of the right triangle is 60 degrees and the other is 90- 60= 30 degrees. If one side of the equilateral triangle (the hypotenuse of the right triangle) has length s, then the side that was cut in half has length s/2. By the Pythagorean theorem, the other leg of the right triangle, an "altitude" of the equilateral triangle, has length c, given by $c^2+ s^2/4= s^2$ so that $c^2= s^2- \frac{s^2}{4}= \frac{3}{4}s$ so $c= \frac{\sqrt{3}}{4}s$.

Sine= opposite side/hypotenuse so now we have $sin(60)= \frac{\frac{s\sqrt{3}}{4}}{s}= \frac{\sqrt{3}}{2}$.
Cosine= near side/hypotenuse so now we have $cos(60)= \frac{\frac{1}{2}s}{s}= \frac{1}{2}$.

Looking at the 30 degree angle just swaps "near side" and "opposite side" so swaps sine and cosine:
$sin(30)= \frac{1}{2}$
$cos(30)= \frac{\sqrt{3}}{2}$.

Of course, we have also the "extreme" cases that are not really triangles: sin(0)= 0, cos(0)= 1, sin(90)= 1, cos(90)= 0.

Other than those, there is no simple way to write sine and cosine of general angles.