0

Guest

Courses New

Given that cos(a-b) + cos(b-c) + cos(c-a) = -3/2. Then find the value of cos(a)+cos(b)+cos(C).

Given that cos(a-b) + cos(b-c) + cos(c-a) = -3/2. Then find the value of cos(a)+cos(b)+cos(C).

Grade:11

1 Answers

Himanshu
103 Points
7 years ago
First of all, multiply 2 and then add 3 on both sides.
Resolve 3 as 1+1+1 on L.H.S and open the identities of cos(x-y).
Replace every 1 as sin2a + cos2a , sin2b + cos2b and sin2c + cos2c respectively.
You will get the two terms (cosa + cosb + cosc)2 + (sina +sinb + sinc)2 = 0
Hence, cosa + cosb + cosc = 0

Think You Can Provide A Better Answer ?

Other Related Questions on Trigonometry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More