proving cofunction relationships?

Question

i missed 2 days of school and now i dont understand the homework i need to 'teach myself'. it has to do with proving cofunction relationships:
First of all, it says in an example on my handout that

sin ( pi / 2 + theta) = [ (a) / (root of all (a^2 + b^2)) ]
and that
cos theta = [ (a) / (root of all (a^2 + b^2)) ]
??? there's a diagram illustrating this on my sheet, with 2 right angle triangles in the first and second quadrants...?
and secondly (the actual question im stuck on...)
it says that using;
the following information, i need to derive a relationship for each of the following and then prove the relationship???

(a) csc (3/2 pi - theta) (b) sec (3/2 pi - theta)
(c) cot (3/2 pi - theta)

i dont understand any of this...so confusing to me...can anyone help please??

(btw if this helps, it says use the relationship in the previous question to answer this, and the prev. question was
sin (270 degrees - theta) / - cos theta

thank you!!!!

Answer 1

(a) csc (3/2 pi - theta)
Θ =1.5π
1/sin(1.5π)

(b) sec (3/2 pi - theta)
Θ =1.5π
1/cos(1.5π)

(c) cot (3/2 pi - theta)
Θ =1.5π
1/tan(1.5π) or = cos(1.5π) / sin(1.5π)

Answer 2

let's apply some values to that diagram you didn't show. you have a right triangle in quad 1, vertical side b, horizontal side a, hypotenuse c =√(a²+b²), and angle between a and c = Θ. for an angle of π/2 + Θ, we set the same triangle in quad 2 with the a side running up the y (π/2) axis, the b side running left from there into quad 2. drop a perpendicular from the terminal end of c down to the negative x axis. the foot of this line is -b, the height is a, and the reference angle is the complement of Θ, which is π/2 - Θ. using the diagram now, see that sin Θ = b/c, cos Θ = a/c, sin(π/2 + Θ) = a/c, cos(π/2 + Θ) = -b/c, and so on. it's all about how the reference triangle to (π/2 + Θ) relates to that congruent triangle back in quad 1.

so to work with (3π/2 - Θ), draw a triangle congruent to the one in quad 1 in quad 3 with the a side running down the negative y axis, the b side running left from there into quad 3. let me use Φ for (3π/2 + Θ) just to make it easier to type. then based on the triangle, sin Φ = -a/c, cos Φ = -b/c, tan Φ = -b/-a = b/a, and then what you're after, csc Φ = -c/a [remember c = √(a²+b²)], sec Φ = -c/b, cot Φ = -a/-b = a/b

it's all about the basic triangle for Θ in quad 1 repositioned into the other quadrants for the other angles. it's all about VISUALIZING it, which is great if you're a visual thinker, not so much if you're not.