*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 + \({\pi \over 2}\)* = (

= - *\({60 \over 169}\)* - *\({60 \over 169}\)*

= - *\({120 \over 169}\)*

However, the answer in the textbook is - *\({12 \over 13}\) . *What did I do wrong?

Guest Feb 22, 2019

#2

#6**+1 **

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?

ElectricPavlov
Feb 22, 2019