SOLUTION: if sec(t) = -5 and the terminal point of t is in quadrant iii, use the identity sin^2(t/2)=1-cos(t)/2 to find the exact value of (sin(t/2)+cos(t))

Algebra ->  Trigonometry-basics -> SOLUTION: if sec(t) = -5 and the terminal point of t is in quadrant iii, use the identity sin^2(t/2)=1-cos(t)/2 to find the exact value of (sin(t/2)+cos(t))      Log On


   



Question 1158051: if sec(t) = -5 and the terminal point of t is in quadrant iii, use the identity sin^2(t/2)=1-cos(t)/2 to find the exact value of (sin(t/2)+cos(t))
Answer by KMST(5377) About Me  (Show Source):
You can put this solution on YOUR website!
I expect the identity was meant to be sin%5E2%28t%2F2%29=%281-cos%28t%29%29%2F2 .
What was written in the question was sin%5E2%28t%2F2%29=1-cos%28t%29%2F2 , which is not an identity.
IMPORTANT NOTE:
On paper, you can draw a long horizontal line under 1-cos%28t%29 and write a 2 under that line.
That long horizontal line implies invisible brackets wrapping together what is above the line,
and wrapping together what is below. Everyone understands that.
When typing, or entering formulas and numbers into a calculator, those brackets need to be typed in or keyed in.
If you want to calculate %28101-2%29%2F%28101-1%29=99%2F100=0.99 ,
and key in "101-2/101-1" into a calculators you would most likely get as a result
101-2%2F101-1=101-1-2%2F101=100-2%2F101=%2299.980198019801...%22 , which is not what you would expect.
I was not expecting anything, you are outsourcing problem solving to a gadget that does not understand what you want, and does just what you asked.

BACK TO THE QUESTION:
Quadrant III means 270%5Eo%3Ct%3C180%5Eo , so that 135%5Eo%3Ct%2F2%3C90%5Eo .
That means the terminal side of angle t%2F2 is in quadrant II,
and sin%28t%2F2%29%3E0 .
sec%28t%29=1%2Fcos%28t%29 is another trigonometric identity,
and from sec%28t%29=-5 , we know that 1%2Fcos%28t%29=-5 --> highlight%28cos%28t%29=-1%2F5%29
Substituting -1%2F5 for cos%28t%29 in the identity sin%5E2%28t%2F2%29=%281-cos%28t%29%29%2F2 , we get
sin%5E2%28t%2F2%29%22=%22%281-%28-1%2F5%29%29%2F2 --> sin%5E2%28t%2F2%29%29%22=%22%281%2B1%2F5%29%2F2 --> sin%5E2%28t%2F2%29%29%22=%22%286%2F5%29%2F2 --> sin%5E2%28t%2F2%29%29%22=%226%2F10 --> sin%5E2%28t%2F2%29%29%22=%223%2F5 --> highlight%28sin%28t%2F2%29=sqrt%283%2F5%29=sqrt%2815%29%2F5%29
Then, highlight%28sin%28t%2F2%29%2Bcos%28t%29=%28sqrt%2815%29-1%29%2F5%29