Question: Prove the following trigonometric identity: $\dfrac{sin 3A}{sinA} - \dfrac{cos 3A}{cosA}$ = 2.

Solution:


You can also solve the above question by the following process using the formula of compound angles.

Solution:
Given,

LHS
= $\dfrac{sin 3A}{sinA} - \dfrac{cos 3A}{cosA}$

= $\dfrac{sin 3A.cosA - cos3A.sinA}{sinAcosA}$

= $\dfrac{sin(3A -A)}{sinAcosA}$

= $\dfrac{sin2A}{sinAcosA}$

= $\dfrac{2sin2A}{2sinAcosA}$

= $\dfrac{2sin2A}{sin2A}$

= 2
RHS


Related Notes And Solutions:

Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles

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