Prove that for any two angles A and B, cos(A-B)= cosA.cosB+sinA.sinB
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Answer:
Step-by-step explanation:
Let angle A = 60° and B = 30°.
L.H.S. = cos ( A - B )
= cos ( 60° - 30°)
= cos 30°
= √3 / 2
R.H.S. = cos A. cos B + sin A. sin B
= cos 60°.cos 30° + sin 60°. sin 30°
= 1 / 2 × √3 / 2 + √3 / 2 × 1 / 2
= √3 / 4 + √3 / 4
= ( √3 + √3 ) / 4
= 2√3 / 4
= √3 / 2
∴ L.H.S. = R.H.S.
Hence, it is proved.
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