Question

if A+B+C= pie,then prove cos2A-cos2B-cos2C=-1+4 cosA sinB sinC

Answer 1

plsss hlp

Answer 2

i want to help but i really dont know the ans.

Answer 3

For answer visit:-

Answer 4

http://openstudy.com/study#/updates/5174d0c4e4b0be9997dfea32

Answer 5

bt it is another question

Answer 6

OK

Answer 7

yeh yeh

Answer 8

\[A+B+C= \pi \rightarrow A+B = \pi -C
\[\sin (A+B)= \sin(\pi-C)= \sin C \]
LHS = cos2A-cos2B-cos2C
\[= -2 \sin \frac{2A+2B}{2} \sin \frac{2A-2B}{2}-\cos2C\]
\[= -2 \sin (A+B) \sin (A-B)-\cos2C\]
\[= -2 \sin (\pi -C) \sin (A-B)-\cos2C\]
\[= -2 \sin(C) \sin (A-B)-(1-2\sin^2C) = -2 \sin(C) \sin (A-B)-1+2\sin^2C \]
\[= 2\sin^2C-2 \sin(C) \sin (A-B)-1 =2sinC [sinC- \sin (A-B)]-1 \]
\[= 2sinC[\sin[\pi-(A+B)]- \sin (A-B)]-1 =2sinC [\sin(A+B)- \sin (A-B)]-1\]
\[=2sinC [2\cos \frac{(A+B)+(A-B)}{2} \sin \frac{(A+B)-(A-B)}{2}-1\]
\[=2sinC [2\cos \frac{A+B+A-B}{2} \sin \frac{A+B-A+B}{2}-1\]
\[=4sinC \cos \frac{2A}{2} \sin \frac{2B}{2}-1\]
\[=4sinC \cos A \sin B-1\]
\[=-1+4sinC \cos A \sin B\]
\[=-1+4 \cos A \sin B \sin C \]
=RHS
Thus LHS=RHS


@arkababu

Answer 9

@arkababu Your problem has been proved. Please check.

Answer 10

thnkss @dpasingh a lot

Answer 11

i want to help but i really dont know the ans.