# Exact Values with the different trig functions

• May 27th 2009, 05:13 PM
hazel31304trighelp
Exact Values with the different trig functions
okay so im having problems with this one specific math problem:

sin 30 cos 60 + sin 60 cos 30

I know that sin = y/r and that cos= x/r but what exactly does radius (r) equal??? doesnt it equal 1?

its a simple question, but its bugging me because i can't seem to get the right answer, which is 1.

ANY HELP??? thanks!!!!
• May 27th 2009, 05:55 PM
skeeter
Quote:

Originally Posted by hazel31304trighelp
okay so im having problems with this one specific math problem:

sin 30 cos 60 + sin 60 cos 30

I know that sin = y/r and that cos= x/r but what exactly does radius (r) equal??? doesnt it equal 1?

its a simple question, but its bugging me because i can't seem to get the right answer, which is 1.

ANY HELP??? thanks!!!!

here's a big hint for you ...

$\textcolor{red}{\sin(a)\cos(b) + \sin(b)\cos(a) = \sin(a+b)}
$
• May 27th 2009, 06:12 PM
matheagle
You can just plug in these values into any basic calculator.
I use a \$10 TI-30, which suits all my needs.
• May 28th 2009, 12:18 AM
Ife
Quote:

Originally Posted by hazel31304trighelp
okay so im having problems with this one specific math problem:

sin 30 cos 60 + sin 60 cos 30

I know that sin = y/r and that cos= x/r but what exactly does radius (r) equal??? doesnt it equal 1?

its a simple question, but its bugging me because i can't seem to get the right answer, which is 1.

ANY HELP??? thanks!!!!

I will try to answer this as best as possible. Think of an equilateral triangle, with length of side equal 1. If this was bisected, we would get 2 right triangles, with hypothenuse of 1, and with interior angles 60, and 30. Let this be the basis of the calculation of the sine and cos exact values in a unit circle. We know that a unit circle has a radius of 1. So if one of those right triangles was at the centre of a unit circle, the hypothenuse would be equal to the radius, and would have a length of 1. Sketch it out on paper. At 60 degrees, the length of the adjacent leg of the triangle would be 1/2. (remember we bisected the the triangle of side = 1, so the length of side is now 1/2. The third side (opposite the angle) would be root 3/2. Use pythagoras' theorem to verify. 1 sq = 1/2 sq + side sq. Therefore: 1-1/4 = side sq, = 3/4, therefore root of 3 / 2 is the length of the 3 rd arm. Since Sin x = y/r, when x is 60, sin 60 = (root 3 / 2 ) /1. cos 60 = (1/2)/1. And for the 30 degrees angle, just rearrange ur triangle, so the y value would be 1/2, and the x value would be root 3/2. If you need more help, feel free to ask. I hope you follow
• May 28th 2009, 12:27 AM
matheagle
Using...

Quote:

Originally Posted by skeeter
here's a big hint for you ...

$\textcolor{red}{\sin(a)\cos(b) + \sin(b)\cos(a) = \sin(a+b)}
$

You should get $\sin(90)=1$ assuming you are talking degrees here.
Or just use your calculator, but make sure it's set to degrees and not radians.