Prove that Sin3A-SinA/Cos3A+Cos5A=Cot2A
Answers
(−3+5−7)/(−3−5+7) = cot 2A
Step-by-step explanation:
Note: There is an error in the question. The question should actually be: (−3+5−7)/(−3−5+7)= cot 2A
To Prove: (−3+5−7)/(−3−5+7) = cot 2A
Proof:
Trigonometric identities:
Sin A - SinB = 2 Cos A+B/2 Sin A-B/2
Sin A + Sin B = 2 Sin A+B/2 Cos A-B/2
Cos A + Cos B = 2 CosA+B/2 CosA-B/2
Cos A - Cos B = -2 Sin A+B/2. Sin A-B/2
Sin (A+B) = SinA.CosB + CosA.SinB
Cos (A+B) = CosA.CosB - SinA.SinB
LHS = (−3+5−7)/(−3−5+7)
= (Sin5A+SinA) -(Sin7A +Sin3A) / (Cos7A+CosA) -(Cos5A +Cos3A)
= (2.Sin 5A+A/2 .Cos5A-A/2 - 2Sin&A+3A/2.Cos7A-3A/2 ) / (2.Cos7A+A/2. Cos7A-A/2 - 2Cos5A+3A/2.Cos5A-3A/2)
= 2Sin3A.Cos2A -2Sin5A.Cos2A / 2Cos4A.Cos3A - Cos4A.CosA
= 2Cos2A (Sin3A - Sin 5A) / 2Cos2A(Cos3A-CosA)
= Cos2A (2Cos5A+3A/2.Sin5A-3A/2) / Cos4A (2SinA+3A/2.Sin3A-A/2)
= Cos2A. Cos4A. SinA / Cos4A.Sin2A. SinA
= Cos2A / Sin2A
= Cot 2A
= RHS
Hence proved.