# Finding the exact vale of 6 Trig Functions For 90 degrees.

• July 4th 2008, 12:25 PM
Dave1234
Finding the exact vale of 6 Trig Functions For 90 degrees.
i know how to do the ones like 30 45 and 60 but i just dont no where to go with this one.
• July 4th 2008, 12:28 PM
Jhevon
Quote:

Originally Posted by Dave1234
i know how to do the ones like 30 45 and 60 but i just dont no where to go with this one.

these you can know from looking at the graphs of the trig functions. if you can figure out the values for sine and cosine, you can figure out the rest. so, what is sin(90) and cos(90)? look at the graphs. the angles are on the x-axis and the values of the functions are on the y-axis.

then, you know that:

tan(90) = sin(90)/cos(90)

sec(90) = 1/cos(90)

cosec(90) = 1/sin(90)

cot(90) = 1/tan(90) = cos(90)/sin(90)
• July 4th 2008, 12:38 PM
Chop Suey
Indeed, like Jhevon said, you need to look at the graph. The unit circle ( $x^2 + y^2 = r^2$) is used to get the values of trigonometric ratios of the angles 0, 90, 270, 360.

Dave, I want you to draw a circle with a radius of 1 on the graph, and observe. Plot the primary points: (1, 0), (0, 1), (-1, 0), (0,-1). Then jooin these 4 points to form a circle. Remember that the points are actually P(cos(x), sin(x)). What do you see at 0 degrees? The points are (1, 0). So, cos (0)=1, and sin(0) =0. Now, what do you see at 90 degrees?