Advertisements
Advertisements
Question
Prove that:
(cos α – cos β)2 + (sin α – sin β)2 = 4 sin2 `((alpha - beta)/2)`
Advertisements
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:
sin 50° + sin 10° = cos 20°
Prove that:
sin 40° + sin 20° = cos 10°
Prove that:
sin 50° − sin 70° + sin 10° = 0
Prove that:
cos 20° + cos 100° + cos 140° = 0
Prove that:
\[\sin\frac{5\pi}{18} - \cos\frac{4\pi}{9} = \sqrt{3} \sin\frac{\pi}{9}\]
Prove that:
sin 3A + sin 2A − sin A = 4 sin A cos \[\frac{A}{2}\] \[\frac{3A}{2}\]
Prove that:
Prove that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), then write the value of tan A tan B tan C.
Express the following as the product of sine and cosine.
sin 6θ – sin 2θ
