Advertisements
Advertisements
Question
Prove that:
(cos α – cos β)2 + (sin α – sin β)2 = 4 sin2 `((alpha - beta)/2)`
Solution
LHS = (cos α – cos β)2 + (sin α – sin β)2
`= (- 2 sin (alpha + beta)/2 sin (alpha - beta)/2)^2 + (2 cos (alpha + beta)/2 sin (alpha - beta)/2)^2`
`= 4 sin^2 (alpha + beta)/2 sin^2 (alpha - beta)/2 + 4 cos^2 (alpha + beta)/2 sin^2 (alpha - beta)/2`
`= 4 sin^2 (alpha - beta)/2 [sin^2 (alpha + beta)/2 + cos^2 (alpha + beta)/2]`
`= 4 sin^2 (alpha - beta)/2` = RHS
APPEARS IN
RELATED QUESTIONS
Prove that:
Prove that:
cos 100° + cos 20° = cos 40°
Prove that:
Prove that:
cos 40° + cos 80° + cos 160° + cos 240° =
If sin 2 θ + sin 2 ϕ = \[\frac{1}{2}\] and cos 2 θ + cos 2 ϕ = \[\frac{3}{2}\], then cos2 (θ − ϕ) =
cos 35° + cos 85° + cos 155° =
Express the following as the product of sine and cosine.
cos 2A + cos 4A
Prove that:
sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0
Evaluate:
sin 50° – sin 70° + sin 10°
