Terminal Arm Point P: Finding sin, cos, and tan for Θ in Standard Position

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SUMMARY

The discussion focuses on calculating the trigonometric values of sinΘ, cosΘ, and tanΘ for a point P with coordinates (-2, 6) on a terminal arm in standard position. The correct calculation of r, using the formula r = √(x² + y²), yields r = √38. The user initially miscalculated sinΘ as 6/√38 but later corrected it to 3/√10 after realizing the error in squaring the x-coordinate. The final values derived from the coordinates are sinΘ = 3/√10, cosΘ = -2/√10, and tanΘ = -3/1.

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Homework Statement



The coordinates of a point P on a terminal arm of an <Θ (theta) in standard position are given, where 0<Θ<2pi. Determine the exact values of sinΘ, cos Θ and tan Θ

2h) (-2,6)



Homework Equations



r= X +y


The Attempt at a Solution



So, first i found r by using r^2 = x^2 + y^2

and got r = √ 38

Since √ 38 cannot be factored to any lower form, i left that as the r.

Going to sin, I did op/hypotenuse

and got 6/√ 38 . However, this seems to be completely wrong as the answer is somehow 3/√ 10

I just don't see how that possible, anyone?
 
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You forgot to square the x coordinate when you calculated r.
 
oops, thanks sir.
 

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