I'm going to explain the relationships between everything using a right triangle. This will be the easiest way to explain everything.
Let's remember the trig relations with respect to #alpha#. I have the corresponding reciprocals on the same line: sine and cosecant, cosine and secant, tangent and cotangent.
#sin alpha=a/c# AND #csc alpha =c/a#
#cos alpha=b/c# AND #sec alpha=c/b#
#tan alpha=a/b# AND #cot alpha=b/a#
Now that we figured that out, let's do the same for #beta#.
#sin beta=b/c# AND #csc beta=c/b#
#cos beta=a/c# AND #sec beta=c/a#
#tan beta=b/a# AND #cot beta=a/b#
Also, remember that the angles #alpha# and #beta# add up to #90^o#, so you get #alpha+beta=90# or #beta=90-alpha#. If you were to convert this to radians, it would be #alpha+beta=pi/2# or #beta=pi/2-alpha#.
This will be really helpful when explaining the phase shift. Phase shift is the change in the x value of the function. Knowing these two things, let's determine the trig equivalents between #alpha# and #beta#
#sin alpha=cos beta# AND #csc alpha = sec beta#
#cos alpha=sin beta# AND #sec alpha = csc beta#
#tan alpha=cot beta# AND #cot alpha = tan beta#
Let's talk about phase shifts and equivalent functions. Why does #sec(90^o-alpha)=csc(alpha)#? Remember when I said that about the angles adding up to #90^o#? This is where that comes into play. Let's see why. remember to substitute #beta# for #90-alpha# and then find the equivalent function from what I listed.
#sec(90-alpha)=sec(beta)=csc(alpha)# (answer to #1)
Let's try with radians this time.
#sin(pi/2-alpha)=sin(beta)=cos(alpha)# (answer to #3... cosine is the equivalent)
If you were to graph #sec(90-alpha)# and #csc alpha#, you will get overlapping graphs. If you were to graph #sin(pi/2-alpha)# and #cos alpha# Changing gears now. Let's talk about the unit circle.
Let's decipher this. The unit circle will always be 1 unit in radius from the origin of the cartesian plane. The angle of measure starts at the x-axis in the positive direction (1,0) and goes around counterclockwise for positive degrees/radians.
What happens when you have negative radians or degrees? You still start at the x-axis in the positive direction, but you go clockwise for negative degrees/radians.
For every #±180^o# or #±pi#, you go around half the circle. For every #±360^o# or #±2pi#, you make one full circle.
The number you see in parentheses are your cosine (x-value) and sine (y-value).
cos = x AND sec = #1/cos#
sin = y AND csc = #1/sin#
tan = #sin/cos# = #y/x# AND cot = #1/tan#
Your cosine is your x-value, and your sine is your y-value. Since #tan theta=(sin theta)/(cos theta)#, the slope of the line is your tangent. Secant is the reciprocal of the x-axis. Cosecant is the reciprocal of the y-axis. Cotangent is the reciprocal of the slope. The numbers inside the circle are your angles.
What are the coordinates for #-(9pi)/2#? Let's change the radians represent a proper fraction.
#-(9pi)/2=-4pi/2=-4pi-pi/2#
This means that we make two full clockwise turns of the circle (#4/2=2#) and then make one quarter clockwise turn of the circle. This is equivalent to #(3pi/2)#, so the answer is (0,-1). (answer to #2)
I'm sorry for the long explanation, but I hope this helps you!
