Prove that (Cos9A – Cos5A)/(Sin17A – Sin3A) = – Sin2A/Cos10A

Let’s simplify left hand side of this equation (Cos9A – Cos5A)/(Sin17A – Sin3A) = – Sin2A/Cos10A

Simplifying Cos9A – Cos5A

Using formulas CosC – CosD = 2 Sin(C + D)/2 Sin(D – C)/2

Replacing
C = 9A
D = 5A

(C + D)/2 = (9A + 5A)/2 = 14A/2 = 7A

(D – C)/2 = (5A – 9A)/2 = – 4A/2 = – 2A

Therefore CosC – CosD = 2 Sin(C + D)/2 Sin(D – C)/2 becomes

Cos9A – Cos5A = 2 Sin7A Sin(- 2A)

Simplifying Sin17A – Sin3A

Using formulas SinC – SinD = 2 Cos(C + D)/2 Sin(C – D)/2

Replacing
C = 17A
D = 3A

(C + D)/2 = (17A + 3A)/2 = 20A/2 = 10A

(C – D)/2 = (17A – 3A)/2 = 14A/2 = 7A

Sin17A – Sin7A = 2 Cos10A Sin7A

(Cos9A – Cos5A)/(Sin17A – Sin3A)

Cos9A – Cos5A = 2 Sin7A Sin(- 2A)

Sin17A – Sin7A = 2 Cos10A Sin7A

Dividing

(Cos9A – Cos5A)/(Sin17A – Sin7A) = (2 Sin7A Sin(- 2A))/(2 Cos10A Sin7A) = Sin(- 2A)/Cos10A

(Cos9A – Cos5A)/(Sin17A – Sin7A) = Sin(- 2A)/Cos10A

For any value of θ, Sin(- θ) = – Sinθ
Therefore Sin(- 2A) = – Sin2A

(Cos9A – Cos5A)/(Sin17A – Sin7A) = – Sin2A/Cos10A

Therefore value of (Cos9A – Cos5A)/(Sin17A – Sin7A) is – Sin2A/Cos10A

Related Posts

Leave a Reply

Your email address will not be published.