Advertisements
Advertisements
प्रश्न
Prove that:
sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0
Advertisements
उत्तर
Consider sin (A – B) sin C
= (sin A cos B – cos A sin B) sin C
= sin A cos B sin C – cos A sin B sin C …….. (1)
Similarly sin(B – C) sin A = sin B cos C sin A – cos B sin C sin A …….. (2)
[Replace A by B, B by C, C by A in (1)]
and sin(C – A) sin B [Replace A by B, B by C, C by A in (2)]
= sin C cos A sin B – cos C sin A sin B …….. (3)
Adding (1), (2) and (3) we get
sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0
APPEARS IN
संबंधित प्रश्न
Show that :
Prove that:
cos 20° cos 40° cos 80° = \[\frac{1}{8}\]
Express each of the following as the product of sines and cosines:
sin 5x − sin x
Prove that:
sin 50° + sin 10° = cos 20°
Prove that:
cos 20° + cos 100° + cos 140° = 0
Prove that:
Prove that:
If A + B = \[\frac{\pi}{3}\] and cos A + cos B = 1, then find the value of cos \[\frac{A - B}{2}\].
Express the following as the sum or difference of sine or cosine:
cos 7θ sin 3θ
Express the following as the product of sine and cosine.
cos 2θ – cos θ
