Advertisements
Advertisements
Question
Prove that:
sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0
Advertisements
Solution
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
RELATED QUESTIONS
Show that:
sin A sin (B − C) + sin B sin (C − A) + sin C sin (A − B) = 0
Prove that:
sin 50° + sin 10° = cos 20°
Prove that:
Prove that:
cos 40° + cos 80° + cos 160° + cos 240° =
If sin (B + C − A), sin (C + A − B), sin (A + B − C) are in A.P., then cot A, cot B and cot Care in
Express the following as the sum or difference of sine or cosine:
`sin "A"/8 sin (3"A")/8`
Express the following as the product of sine and cosine.
cos 2A + cos 4A
Prove that:
`(cos 7"A" +cos 5"A")/(sin 7"A" −sin 5"A")` = cot A
If secx cos5x + 1 = 0, where 0 < x ≤ `pi/2`, then find the value of x.
