We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

*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