CLEAR
Toprankers-Info

+91 7676564400

Mobile App

Get it on Google Play
Discussions
Select Date
Tags:

Articles

Discussions
Submit Discussion
Vijay Kumar

· started a discussion

· 1 Months ago

PUT Q (Question:110226.0)

let A=B=C=90

Question:

If A + B + C = \(\cfrac{3\pi}{2}\), then the value of cos 2A + cos2B + cos 2C is equal to:

Options:
A) 1 - 4 cos A cos B cos C
B) 1 - 4 sin A sin B sin C
C) 0
D) 1
Solution:
Ans: (b) Given that A + B + C = \(\cfrac{3\pi}{2}\)

\(\therefore\) cos 2A + cos 2B + cos 2C

=(cos 2A + cos 2B) + cos 2C

= 2 cos (A + B) cos (A - B) + cos 2C

= -2 sin C cos (A - B) + (1 - 2 sin2C)

= 1-2 sin C [cos (A-B) + sin C]

= 1 - 2 sin C [cos (A-B) - cos (A+B)]

[∵ sin C = -cos(A + B) by A + B + C = \(\cfrac{3\pi}{2}\)]

= 1 - 2 sin C  [2sinAsin B]

[∵ cos (A-B) - cos (A+B) = 2 sinA sin B]

= 1 - 4 sin A sin B sin C

Knowledge Expert

· commented

· 1 Months ago

Dear Student,
Kindly clarify your doubts.
Keep learning with us
Thank You
Team Toprankers
-logo
We at aim to provide most comprehensive content & test for practice which is carefully divided into various chapters and topics to help you focus on your weak areas.Our Practice mock test gives you unmatched analytics to help you realize your strong areas,time management and improvement areas.Our Short & Crisp notes on each topic will help you to quickly revise the topic along with the tips & tricks of the exams.
More About
-Info
-Info

Mobile App

Get it on Google Play
About Us Campus Ambassador Careers Contact Us Refund Policy

All Rights Reserved Top Rankers

User Policy
Disclaimer
Terms & Conditions