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
If 4 cos2 x° - 1 = 0 and 0 ∠ x° ∠ 90°,
find:(i) x°
(ii) sin2 x° + cos2 x°
(iii) `(1)/(cos^2xx°) – (tan^2 xx°)`
Calculate the value of A, if (tan A - 1) (cosec 3A - 1) = 0
If sin 3A = 1 and 0 < A < 90°, find cos 2A
If sin α + cosβ = 1 and α= 90°, find the value of 'β'.
Solve for 'θ': `sec(θ/2 + 10°) = (2)/sqrt(3)`
Find the value of 'x' in each of the following:
Find x and y, in each of the following figure:
Evaluate the following: `(5sec68°)/("cosec"22°) + (3sin52° sec38°)/(cot51° cot39°)`
Evaluate the following: `(3sin^2 40°)/(4cos^2 50°) - ("cosec"^2 28°)/(4sec^2 62°) + (cos10° cos25° cos45° "cosec"80°)/(2sin15° sin25° sin45° sin65° sec75°)`
If A, B and C are interior angles of ΔABC, prove that sin`(("A" + "B")/2) = cos "C"/(2)`
