3 years ago
Given: sin theta = (15/17) and ( pi/2) < theta < pi
Find the exact value of: cos ( theta/2)
A) (5 sqrt(34)/34) New York City
B) - (5 sqrt(34)/34) Montreal, Canada
C) (3 sqrt(34)/34) Barcelona, Spain
D) - (3 sqrt(34)/34) Nassau, Bahamas
1 Answers
Best Answer
dcodarby3062 Staff answered 3 years ago
dcodarby3062 Staff answered 3 years ago
Step 1
Ans: Given, sin theta = (15/17)
( pi/2) < theta < pi
We have to find exact value of cos ( theta/2)
Since,
sin theta = (Perpendicular/Hypotenuse)
Rightarrow Perpendicular = 15
Hypotenuse = 17
We know that,
Pythagorean theorem
Hypotenuse^(2) = Perpendicular^(2) + Base^(2)
Rightarrow (17)^(2) = (15)^(2) + Base^(2)
Rightarrow Base^(2) = 289 - 225 = 64
Rightarrow Base = 8
Since,
cos theta = (Base/Hypotenuse)
Rightarrow cos theta = (8/17)
because ( pi)[2) < theta < pi (2^nd quadrant)
So,
cos theta = (-8/17)
Now,
We know that
cos^(2) theta = 2 cos^(2) theta - 1
Rightarrow cos^(2) theta = (1 + cos^(2) theta/2)
Rightarrow cos^(2) ( theta/2) = (1 + cos theta/2)
Rightarrow cos ( theta/2) = sqrt( (1 + cos theta/2))
because ( pi/2) < theta < pi Rightarrow ( pi)[4) < ( theta/2) < ( pi/2) (1^(st) quadrant)
Rightarrow cos ( theta/2) = sqrt( (1 + ( (-8/17))(2)) = sqrt( (17 - 8/17 *2)) = sqrt( (9/34)) = (3/ sqrt(34)) = (3/ sqrt(34)) * ( sqrt(34)/ sqrt(34)) = (3 sqrt(34)/34)
Hense, cos ( theta/2) = (3 sqrt(34)/(34))
rightarrow Barselona, Spain