How do you prove that $sin (2 A) = \frac{2 tan (A)}{1 + {tan}^{2} (A)}$ $sin(2A) = (2tan(A))/(1 + tan^2(A))$ ?

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How do you prove that $sin (2 A) = \frac{2 tan (A)}{1 + {tan}^{2} (A)}$ $sin(2A) = (2tan(A))/(1 + tan^2(A))$ ?

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Answer 1 · 2015-04-24T15:07:56

The left member can be writtten in this way:

$(1) = 2 sin A cos A$ $(1)=2sinAcosA$

The right member can be written in this way:

$(2) = \frac{2 tan A}{1 + {tan}^{2} A} = 2 \cdot \frac{\frac{sin A}{cos A}}{1 + \frac{{sin}^{2} A}{{cos}^{2} a}} = 2 \cdot \frac{\frac{sin A}{cos A}}{\frac{{cos}^{2} A + {sin}^{2} A}{{cos}^{2} A}} =$ $(2)=(2tanA)/(1+tan^2A)=2*(sinA/cosA)/(1+sin^2A/cos^2a)=2*(sinA/cosA)/((cos^2A+sin^2A)/cos^2A)=$

$= 2 \cdot \frac{sin A}{cos A} \cdot \frac{{cos}^{2} A}{1} = 2 sin A cos A = (1)$ $=2*sinA/cosA*cos^2A/1=2sinAcosA=(1)$ .

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How do you prove that $sin (2 A) = \frac{2 tan (A)}{1 + {tan}^{2} (A)}$ $sin(2A) = (2tan(A))/(1 + tan^2(A))$ ?

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How do you prove that $sin (2 A) = \frac{2 tan (A)}{1 + {tan}^{2} (A)}$ $sin(2A) = (2tan(A))/(1 + tan^2(A))$ ?