Prove that sin(2A) = 2tan(A) / 1 + tan^2(A)
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Sin (a+b)=SinaCosb+CosaSinb
So Sin(2A)= 2SinACosA = 2sinA/cosA *Cos^A = 2tanA * cos^A = 2 tanA * (cos^A/1)
= 2tanA* [cos^A / (cos^A + sin^A)] =2tanA/ [(cos^A+ sin^A)/ cos^A] = 2tanA/ [ 1+ sin^a/cos^A]
= 2 tanA/ [1 + tan^A]
So Sin(2A)= 2SinACosA = 2sinA/cosA *Cos^A = 2tanA * cos^A = 2 tanA * (cos^A/1)
= 2tanA* [cos^A / (cos^A + sin^A)] =2tanA/ [(cos^A+ sin^A)/ cos^A] = 2tanA/ [ 1+ sin^a/cos^A]
= 2 tanA/ [1 + tan^A]
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RHS = 2tan(A)/(sec^2 (A)) = 2tan(A)cos^2(A) = 2sin(A)cos(A) = sin(2A) = LHS
On a side note, if you're struggling with these trig identities, you should practice many more problems. Trigonometric substitutions are useful for simplifying more challenging calculus problems so it is important that you know your basics well. I wish you all the best!
On a side note, if you're struggling with these trig identities, you should practice many more problems. Trigonometric substitutions are useful for simplifying more challenging calculus problems so it is important that you know your basics well. I wish you all the best!