Question: Prove the following trigonometric identity: \dfrac{sin5A}{sinA} - \dfrac{cos5A}{cosA} = 4 cos2A.
Solution:
Given,
LHS
= \dfrac{sin 5A}{sinA} - \dfrac{cos 5A}{cosA}
= \dfrac{sin 5A.cosA - cos5A.sinA}{sinAcosA}
= \dfrac{sin(5A -A)}{sinAcosA}
= \dfrac{sin4A}{sinAcosA}
= \dfrac{sin(2*2A)}{sinAcosA}
= \dfrac{2sin2A.cos2A}{sinAcosA}
= \dfrac{2 * 2sin2A.cos2A}{2sinAcosA}
= \dfrac{4 sin2A.cos2A}{sin2A}
= 4 cos2A
RHS
Related Notes And Solutions:
Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles
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