If A = B = 60°, verify that: sin(A - B) = sinA cosB - cosA sinB

Sum

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.

shaalaa.com

Trigonometric Equation Problem and Solution

Is there an error in this question or solution?

Chapter 22: Trigonometrical Ratios of Standard Angles - Exercise 27.1

Chapter 22 Trigonometrical Ratios of Standard Angles
Exercise 27.1 | Q 15.2

Use the given figure to find:
(i) tanθ°
(ii) θ°
(iii) sin²θ° - cos²θ°(iv) Use sin θ° to find the value of x.

Find the magnitude of angle A, if 2 sin A cos A - cos A - 2 sin A + 1 = 0

Solve for x : cos `(x/(2)+10°) = (sqrt3)/(2)`

Solve for x : sin² 60° + cos² (3x- 9°) = 1

Evaluate the following: `(sin25° cos43°)/(sin47° cos 65°)`

Evaluate the following: `(sin36°)/(cos54°) + (sec31°)/("cosec"59°)`

Evaluate the following: `(5sec68°)/("cosec"22°) + (3sin52° sec38°)/(cot51° cot39°)`

Evaluate the following: `(sin0° sin35° sin55° sin75°)/(cos22° cos64° cos58° cos90°)`

Evaluate the following: `(2sin25° sin35° sec55° sec65°)/(5tan 29° tan45° tan61°) + (3cos20° cos50° cot70° cot40°)/(5tan20° tan50° sin70° sin40°)`

Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos²θ