A+B+C=180 DEGREES
find value of
Sin2A+Sin2B+6sin2C and
Cos2A+Cos2B+6Cos2C
Asked by Vidushi412
7th October 2018,
2:56 PM
Answered by Expert
Answer:
sin2A + sin2B + 6sin2C
= 2sin(A + B)cos(A - B) + 6(2sinCcosC)
= 2sinC cos(A - B) + 6(2sinC cosC)
= 2sinC [cos (A - B) + 6cosC)]
= 2sinC[cos (A - B) - 6cos(A + B)]
= 2sinC[cos(A - B) - cos(A + B) - 5cos(A + B)]
Continue this further ....
Answered by Expert
10th October 2018,
9:56 AM
Rate this answer
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
You have rated this answer /10
Your answer has been posted successfully!
RELATED STUDY RESOURCES :
Browse free questions and answers by Chapters
- 1 Sequences and Series
- 2 Mathematical Reasoning
- 3 Differential Equations
- 4 Three Dimensional Geometry
- 5 Complex Numbers and Quadratic Equations
- 6 Permutations and Combinations
- 7 Matrices and Determinants
- 8 Vector Algebra
- 9 Trigonometry
- 10 Integral Calculus
- 11 Sets, Relations and Functions
- 12 Limit, Continuity and Differentiability
- 13 Statistics and Probability
- 14 Co-ordinate Geometry
- 15 Mathematical Induction
- 16 Binomial Theorem and its Simple Applications
Trending Tags
Latest Questions
JEE Main Mathematics
Asked by amaladelsingh
28th October 2020,
10:24 PM
CBSE X Biology
Asked by reenacharan2008
28th October 2020,
9:11 PM
