Advertisements
Advertisements
प्रश्न
Prove that:
sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0
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
Prove that:
Express each of the following as the product of sines and cosines:
cos 12x + cos 8x
Prove that:
Prove that:
If (cos α + cos β)2 + (sin α + sin β)2 = \[\lambda \cos^2 \left( \frac{\alpha - \beta}{2} \right)\], write the value of λ.
Express the following as the sum or difference of sine or cosine:
cos(60° + A) sin(120° + A)
Prove that:
2 cos `pi/13` cos \[\frac{9\pi}{13} + \text{cos} \frac{3\pi}{13} + \text{cos} \frac{5\pi}{13}\] = 0
Evaluate:
sin 50° – sin 70° + sin 10°
If cos A + cos B = `1/2` and sin A + sin B = `1/4`, prove that tan `(("A + B")/2) = 1/2`