Question 921332: Suppose sec t = 2 and csc t > 0. Find each of the following:
cos t =
cos (−t) =
cos (π + t) =
sin (−t) =
tan (t + π) =
I think I have this right since sec = 2 it's therefore 2/1 and that's hyp/opp so I use the Pythagorean theorem to find A
A^2 = 2^2 + 1^2
A = √(5)
So I have the opposite, adjacent and hypotenuse. I just don't understand what it's asking with -t? π+t? and t+π
please someone explain this
Thanks!
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! cos t = 1/sec(t)
cos(t) = 1/2
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cos (−t) = cos(t) since cosine is an even function
cos(-t) = 1/2
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cos (π + t) = cos(pi)*cos(t) - sin(pi)*sin(t) ... trig identity
cos (π + t) = -1*cos(t) - 0*sin(t)
cos (π + t) = -cos(t)
cos (π + t) = -(1/2)
cos (π + t) = -1/2
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sin (−t) = -sin(t) since sine is an odd function
sin(-t) = -sqrt(3)/2
Note: your triangle should have 1/2 as the leg and 1 as the hypotenuse. So a = 1/2, c = 1. The missing side is b for now
a^2 + b^2 = c^2
(1/2)^2 + b^2 = 1^2
1/4 + b^2 = 1
b^2 = 1 - 1/4
b^2 = 3/4
b = sqrt(3/4) ... csc(t) > 0, sin(t) > 0
b = sqrt(3)/2
So that's how I got sqrt(3)/2. You can also use the unit circle.
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tan (t + π) = [tan(t) + tan(pi)]/[1 - tan(t)*tan(pi)] ... trig identity
tan (t + π) = [tan(t) + 0]/[1 - tan(t)*0]
tan (t + π) = [tan(t) + 0]/[1 - 0]
tan (t + π) = [tan(t)]/[1]
tan(t + π) = tan(t)
This makes sense because tangent has period of pi. It's going to repeat itself every pi units which is exactly what tan(t + π) = tan(t) tells us.
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Let me know if that helps or not. Thanks.
If you need more help, feel free to email me at jim_thompson5910@hotmail.com
My Website: http://www.freewebs.com/jimthompson5910/home.html
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