Question

1). sin 2A +
cos2A + cos 2B + cos 2 (A - B) + 1 = 4 cos A. cos B. Cos (A - B)
sin 2A + sin 2B + sin 2C - sin 2 (A+B+C) = 4 sin (B+C) sin (C+A​

Answer 1

Answer:

2

Step-by-step explanation:

Here we want +1 and as such we write cos2A=1−2sin

2

A and combine the other two terms.

L.H.S. =1−2sin

2

A+2sin(B+C)sin(C−B) (Note)

=1−2sin

2

A−2sinAsin(B−C) [∵sin(−θ)=−sinθ]

=1−2sinA[sinA+sin(B−C)]

=1−2sinA[sin(B+C)+sin(B−C)]

=1−2sinA(2sinBcosC)

=1−4sinAsinBcosC.