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```Solved Examples on Trigonometric Equations and Identities

Illustration 1: If cos (a + b) = 4/5, sin (a-b) = 5/13 and a and b lie between 0 to π/4, find tan 2a.

Solution: It is given that cos (a + b) = 4/5, sin (a-b) = 5/13

It follows that sin (a + b) = 3/5, cos (a-b) = 12/13

Hence, tan (a+b) = 3/4 and tan (a-b) = 5/12

Hence, this implies

tan [(a+b) + a-b] =[tan (a+b) + tan (a-b)]/ [1 + tan(a+b)tan(a-b)]

= (3/4 + 5/12)/ (1 – 3/4. 5/12)

Hence, tan 2a = 14/12. 48/33 = 56/33.

Illustration 2: If A + B + C = 1800, prove that

sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C.

Solution:We have

sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C

So consider LHS = sin (B + C – A) + sin (C + A – B) + sin (A + B – C)

= sin (π– A – A) + sin (π – B – B) + sin (π– C – C)(since A + B + C = π)

= sin 2A + sin 2B + sin 2C

= 4 sin A sin B sin C

Illustration 3: Solve the equation 5 sinθ – 2 cos2θ – 1 = 0

Solution: Given, 5 sinθ – 2 cos2θ – 1 = 0

or,5 sinθ – 2 (1 – sin2θ) – 1 = 0

or,2 sin2 θ + 5 sin θ – 3 = 0

or,(sin θ + 3) (2 sin θ – 1) = 0

∴sin θ = 3 or, sin θ = 1/2

if sin θ = 3

This is not possible as range of sine is [–1, 1]

If sin θ = 1/2

or, sin θ = sin π/6

⇒ θ = n π + (–1)n π/6

where n = 0, ±1, ±2 ……
```

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