SOLUTION: Very confused here... Suppose csc t = 4 and tan t > 0 sin(t - 6π) = cot(5π-t) = cos(-t) = sin(2t) = cos(t+5π/3) = I am having confusion with a f

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Question 922568: Very confused here...

Suppose csc t = 4 and tan t > 0
sin(t - 6π) =
cot(5π-t) =
cos(-t) =
sin(2t) =
cos(t+5π/3) =

I am having confusion with a few things. I know that I would use a right triangle in the first quadrant to find the missing adjacent angle. However, I posted this question early and was told that instead of using a^2 = c^2+b^2 for this it is a^2 = c^2 - b^2 which produces √15 as opposed to √17. I am even more confused after that because I don't understand what the pieces in the parentheses are asking me to do. Such as t-6π, 5π-t and so on.
Please explain how this works
THANKS

Found 2 solutions by josmiceli, Edwin McCravy:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
I think you're having trouble "reading" these functions
------------------
It helps if you can visualize the plot of and
for example,

is called the phase angle
If you set then you have:


This is a multiple of , and since
, then

So, I have:

Here's the plot:

As you can see, there is no difference between this
function and
Now I can say:



-----------------------
You are correct that the angle whose csc is
and whose tan is > 0 is in the 1st quadrant

If
, then


----------------
Now note that

Hope this helps a little -you just have to keep
testing yourself on what you are vague on



Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
You need to learn some trigonometric identities:

1.  sin(A+B) = sin(A)cos(B)+cos(A)sin(B)
2.  sin(A-B) = sin(A)cos(B)-cos(A)sin(B) 
3.  cos(A+B) = cos(A)cos(B)-sin(A)sin(B)
4.  cos(A-B) = cos(A)cos(B)+sin(A)sin(B)
5.  
5.  
6.  
7.  
8.  
9.   for all integers n
10.  for even integers n 
     for odd integers n  

Suppose csc t = 4 and tan t > 0
4 is positive so csc is positive and tangent is positive,
so t is in quadrant 1.

The cosecant is , so draw a right
triangle containing angle t.  Make the hypotenuse 4 and the 
side opposite angle t be 1.  Then csc(t) will be 4/1 or 4.



Since the hypotenuse = 4 and the opposite side = 1, then 







So we put  on the adjacent side:




sin(t - 6π) =
First use 2 above.
 

Now use 9 and 10 above 




cot(5π-t) =
First use 7 above
 

Now use 4 and 2

Now use 9 and 10 and the right triangle to substitute for all the sines
and cosines



cos(-t) =
Write -t as 0-t and use 4



Now use 9 and 10 and the right triangle to substitute for all the sines
and cosines





sin(2t) =
Write 2t as t+t and use 1:



Use the right triangle above to substitute for the sines and cosines:







cos(t+5π/3) =
Use 8



Use 3



300° is one of the special angles in quadrant IV with reference angle 60°

, 





Edwin

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