As usual, make a sketch of a right-angled triangle and using Pythagoras,
r^2 = 8^2 + 15^2 = 289
r = 17
π<α<3π/2 ----> α is in quadrant III
sin α = -15/17 and cos α = -8/17
recall cos 2A = 1 - 2sin^2 A
thus: cos α = 1 - 2sin^2 (α/2)
-8/17 = 1 - 2sin^2 (α/2)
2sin^2 (α/2) = 1 + 8/17 = 25/17
sin^2 (α/2) = 25/34
sin (α/2) = 5/√34 , because (α/2) is in quadrant II
recall that cos 2A = 2cos^2 A - 1
cos α = 2cos^2 (α/2) - 1
-8/17 = 2cos^2 (α/2) - 1
1 - 8/17 = 2cos^2 (α/2)
9/17 = 2cos^2 (α/2)
9/34 = cos^2 (α/2)
cos (α/2) = -3/√34 , since the cosine is negative in II
since tan (α/2) = sin (α/2) / cos (α/2)
= (5/√34) / (-3/√34)
= - 5/3
Hi Tutors,
I've tried solving the problem below, but I keep getting it wrong. I've solved 3 similar questions thanks to Reiny and Steve, but this one, in particular, is giving me problems.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Use the given information to determine the values of sine α/2, cos α/2, and tan α/2.
tanα=15/8;π<α<3π/2
Please show how you got the answers so I can study it step by step.
2 answers
Thank you Reiny
2 answers