Advertisements
Advertisements
Question
Prove that cot 4x (sin 5x + sin 3x) = cot x(sin 5x - sin 3x).
Solution
LHS = cot 4x (sin 5x + sin 3x)
`= cot 4x xx 2 sin ((5x + 3x)/2) cos ((5x - 3x)/2)`
`= 2 (cos 4x)/(sin 4x) xx sin 4x xx cos x` = 2 cos 4x cos x
RHS = cot x (sin 5x - sin 3x)
`= cot x xx 2 sin ((5x - 3x)/2) * cos ((5x + 3x)/2)`
`[sin "A" - sin "B" = 2 sin (("A - B")/2) cos(("A + B")/2)]`
`= (cos x)/cancel(sin x) xx 2 cancel (sin x) cos 4x`
= 2 cos x cos 4x
LHS = RHS.
Hence proved.
APPEARS IN
RELATED QUESTIONS
Find the value of the following:
cot 75°
If sin A = `3/5`, 0 < A < `pi/2` and cos B = `(-12)/13`, π < B < `(3pi)/2`, find the values of the following:
- cos(A + B)
- sin(A – B)
- tan(A – B)
If sin A = `12/13`, find sin 3A.
Prove that `2 sin^2 (3pi)/4 + 2 cos^2 pi/4 + 2 sec^2 pi/3` = 10
Find the value of tan 15°.
If cos (α + β) = `4/5` and sin (α - β) = `5/13` where (α + β) and (α - β) are acute, then find tan 2α.
The value of sin 15° is:
The value of sin 28° cos 17° + cos 28° sin 17°
The value of sec A sin(270° + A) is:
The value of 1 – 2 sin2 45° is: