Angle θ is in quadrant II and sin θ = \({5 \over 13}\). Determine the exact value for sin (θ + \({\pi \over 2}\)).
So I know this is on the triangle 5 - 12 - 13; here's what I've done:
sin θ + \({\pi \over 2}\) = (\({5 \over 13}\))(-\({12 \over 13}\)) + (-\({12 \over 13}\))(\({5 \over 13}\))
= - \({60 \over 169}\) - \({60 \over 169}\)
= - \({120 \over 169}\)
However, the answer in the textbook is - \({12 \over 13}\) . What did I do wrong?
No....I think your APPROACH was OK....the execution of it was wrong
remember cos pi/2 = 0 sooooo sin (x+ pi/2) = sin x cos pi/2 + cos x sin pi/2 cos pi/2 = 0 and sin pi/2 = 1
sin ( x + pi/2) = 0 + cos x SEE?