Question

sin 5pi/8 - cos 4pi/9 = root3 sin pi/9

Answer 1

ANSWER:

Let us consider the LHS sin 5π/18 – cos 4π/9 = sin 5π/18 – sin (π/2 – 4π/9) (since, cos A = sin (90° – A)) = sin 5π/18 – sin (9π – 8π)/18 = sin 5π/18 – sin π/18 On using the formula, sin A – sin B = 2 cos (A + B)/2 sin (A - B)/2 = 2 cos (6π/36) sin (4π/36) = 2 cos π/6 sin π/9 = 2 cos 30° sin π/9 = 2 × √3/2 × sin π/9 = √3 sin π/9 = RHS Thus proved.

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

Let us consider the LHS sin 5π/18 – cos 4π/9

= sin 5π/18 – sin (π/2 – 4π/9) (since, cos A = sin (90° – A))

= sin 5π/18 – sin (9π – 8π)/18

= sin 5π/18 – sin π/18 On using the formula, sin A – sin B

= 2 cos (A + B)/2 sin (A - B)/2

= 2 cos (6π/36) sin (4π/36) = 2 cos π/6 sin π/9

= 2 cos 30° sin π/9

= 2 × √3/2 × sin π/9

= √3 sin π/9

= RHS Thus proved.