Posted by **Josh** on Monday, December 6, 2010 at 10:19am.

Evaluate all six trigonometric functions at t, where the given point lies on the terminal side of an angle of t radians in standard position. (40/41, -9/41)

- Precalculus Someone please help -
**bobpursley**, Monday, December 6, 2010 at 10:37am
draw the triangle, but first, scale it (that wont change the angle). (40,-9) works. That makes the hypotenuse sqrt(40^2+9^2)-=41

so you have a triangle, in the fourth quadrant sides 40, -9, with a hypotenuse of 41

sin= -9/41

cos= 40/41

You take it from here.

- Precalculus Someone please help -
**Reiny**, Monday, December 6, 2010 at 10:39am
You know you are in quadrant IV

by CAST, the cosine, (and the secant), are positive in that quadrant. All other trig ratios are negative.

You also know you are dealing with a 9-40-41 right-angled triangle, since 9^2 + 40^2 = 41^2.

so x=40, y = -9 , and r = 41

Now you can state the other 5 trig ratios.

e.g. tan t = -9/41 (because tanĂ˜ = y/x, etc)

- oops typo - Precalculus Someone please help -
**Reiny**, Monday, December 6, 2010 at 10:41am
e.g. tan t = -9/40

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