how to find all 6 trigonomentric functions problem??

Question

it says to find all 6 trigonometric functions for this...

pheta= tan^-1(-5/12)


please help!! thank you

Answer 1

θ = arctan(-5/12)

Obviously tan θ = -5/12 and cot θ = 1/tan θ = -12/5.

Now √(5^2 + 12^2) = 13, so we're looking at a 5-12-13 right triangle with θ opposite the side with length 5. However, depending on your range for arctan the answers will vary.

For arctan θ between -π/2 and π/2 (the usual case):
cos θ is positive and sin θ is negative (since tan θ is negative), so we have
cos θ = 12/13, sec θ = 13/12
sin θ = -5/13, cosec θ = -13/5

For arctan θ between 0 and π:
sin θ is positive and cos θ is negative (since tan θ is negative), so we have
cos θ = -12/13, sec θ = -13/12
sin θ = 5/13, cosec θ = 13/5.

Answer 2

Katie: One trick: draw a picture. You have a negative sign involved; but is unclear that the theta value is in the 4th quadrant or in the 2nd quadrant. I will solve for both cases, so you can get a clear idea as of the procedure involved.

4th quadrant (x= 5, y = -12) r
Recall definition of tangent (theta) = opposite/adjacent
and that the hypotenuse of a right angle = sqrt (x^2 + y^2)
H = sqrt(25+144) = sqrt(169) = 13
so you have everything you need to solve for the six trigonometric functions.

sin(theta) = 5/13
cos(theta)= -12/13
tan(theta) = 5/-12
csc(theta) = 1/sin(theta) = 13/5
sec(theta) = 1/cos(theta) = -13/12
cot(theta) = 1/tan(theta) = -12/5


For the second quadrant case:

sin(theta) = 5/13
cos(theta) = -12/13
tan(theta) = 5/-12
csc(theta) = 1/sin(theta) = 13/5
sec(theta) = 1/cos(theta) = -13/12
cot(theta) = 1/tan(theta) = -12/5

Be sure to track the definition of theta; sometimes, like in the second quadrant case, the angle you are looking for is (180-theta).

Good luck!

Answer 3

I take it "pheta" (which sounds like a type of cheese) is suppose to be "theta".

If you know tan(θ), then you know cot(θ) because this would just be the reciprocal. Use tan^2(θ) + 1 = sec^2(θ) to find sec(θ). From there you can find cos(θ) as 1/sec(θ), then sin(θ) from sin^2(θ) + cos^2(θ) = 1 (or from tan(θ) = sin(θ)/cos(θ)), then finally csc(θ) as 1/sin(θ).