Is tan(θ)=(sin(2θ))/(1+cos(2θ)) ?

The answer to whether tan(θ)=(sin(2θ))/(1+cos(2θ)) is True

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prove tan(θ)=(sin(2θ))/(1+cos(2θ))

prove

tan (θ) = \frac{sin ( 2 θ )}{1 + cos ( 2 θ )}

True

Solution steps

tan (θ) = \frac{sin ( 2 θ )}{1 + cos ( 2 θ )}

Manipulating right side

\frac{sin ( 2 θ )}{1 + cos ( 2 θ )}

Rewrite using trig identities

\frac{sin ( 2 θ )}{1 + cos ( 2 θ )}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= \frac{2 sin ( θ ) cos ( θ )}{1 + cos ( 2 θ )}

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

= \frac{2 sin ( θ ) cos ( θ )}{1 + 2 cos ^{2} ( θ ) - 1}

Simplify

\frac{2 sin ( θ ) cos ( θ )}{1 + 2 cos ^{2} ( θ ) - 1} : \frac{sin ( θ )}{cos ( θ )}

\frac{2 sin ( θ ) cos ( θ )}{1 + 2 cos ^{2} ( θ ) - 1}

1 + 2 cos^{2} (θ) - 1 = 2 cos^{2} (θ)

1 + 2 cos^{2} (θ) - 1

Group like terms

= 2 cos^{2} (θ) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (θ)

= \frac{2 sin ( θ ) cos ( θ )}{2 cos ^{2} ( θ )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{sin ( θ ) cos ( θ )}{cos ^{2} ( θ )}

Cancel the common factor:

cos (θ)

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{cot ^{2} ( b ) - 1}{2 cot ( b )} = cot (2 b)

\frac{cos ( x ) + sin ( x )}{sin ( x )} = 1 + cot (x)

sin (\frac{π}{3}) = cos (\frac{π}{6})

cos^{2} (t) = \frac{( 1 + cos ( 2 t ) )}{2}

cos (- \frac{π}{6}) = sin (\frac{2 π}{3})

Is tan(θ)=(sin(2θ))/(1+cos(2θ)) ?
The answer to whether tan(θ)=(sin(2θ))/(1+cos(2θ)) is True