Advertisements
Advertisements
Question
If A = B = 60°, verify that: sin(A - B) = sinA cosB - cosA sinB
Advertisements
Solution
sin(A - B) = sinA cosB - cosA sinB
L.H.S. :
sin(A - B) = sin(60°- 60°) = sin0° = 0
R.H.S. :
sinA cosB - cosA sinB
= sin60° cos60° - cos60° sin60°
= `sqrt(3)/(2) xx (1)/(2) - (1)/(2) xx sqrt(3)/(2)`
= `sqrt(3)/(4) - sqrt(3)/(4)`
= 0
L.H.S = R.H.S.
Therefore,
sin(A - B) = sinA cosB - cosA sinB.
APPEARS IN
RELATED QUESTIONS
Solve the following equation for A, if sec 2A = 2
Solve for x : 2 cos (3x − 15°) = 1
Evaluate the following: `((sin3θ - 2sin4θ))/((cos3θ - 2cos4θ))` when 2θ = 30°
In a right triangle ABC, right angled at C, if ∠B = 60° and AB = 15units, find the remaining angles and sides.
Find the length of AD. Given: ∠ABC = 60°, ∠DBC = 45° and BC = 24 cm.
Find x and y, in each of the following figure:
Evaluate the following: `(sec34°)/("cosec"56°)`
Evaluate the following: `(sin36°)/(cos54°) + (sec31°)/("cosec"59°)`
If A, B and C are interior angles of ΔABC, prove that sin`(("A" + "B")/2) = cos "C"/(2)`
Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos2θ
