Is (1-cos(2θ)+sin(2θ))/(1+cos(2θ)+sin(2θ))=tan(θ) ?

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

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Popular Trigonometry >

prove (1-cos(2θ)+sin(2θ))/(1+cos(2θ)+sin(2θ))=tan(θ)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Multiply by

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

= \frac{\frac{( 1 - c o s ( 2 θ ) ) ( 1 + c o s ( 2 θ ) )}{1 + c o s ( 2 θ )} + sin ( 2 θ )}{1 + cos ( 2 θ ) + sin ( 2 θ )}

Apply Difference of Two Squares Formula:

2 θ^{2} - y^{2} = (2 θ + y) (2 θ - y)

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

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

Use the Pythagorean identity:

1 = cos^{2} (2 θ) + sin^{2} (2 θ)

1 - cos^{2} (2 θ) = sin^{2} (2 θ)

= \frac{\frac{s i n ^{2} ( 2 θ )}{1 + c o s ( 2 θ )} + sin ( 2 θ )}{1 + cos ( 2 θ ) + sin ( 2 θ )}

Simplify

\frac{\frac{s i n ^{2} ( 2 θ )}{1 + c o s ( 2 θ )} + sin ( 2 θ )}{1 + cos ( 2 θ ) + sin ( 2 θ )} : \frac{sin ( 2 θ )}{1 + cos ( 2 θ )}

\frac{\frac{s i n ^{2} ( 2 θ )}{1 + c o s ( 2 θ )} + sin ( 2 θ )}{1 + cos ( 2 θ ) + sin ( 2 θ )}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

= \frac{\frac{s i n ^{2} ( 2 θ ) + s i n ( 2 θ ) ( c o s ( 2 θ ) + 1 )}{1 + c o s ( 2 θ )}}{1 + cos ( 2 θ ) + sin ( 2 θ )}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

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

Factor

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

sin^{2} (2 θ) + sin (2 θ) (1 + cos (2 θ))

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

sin^{2} (2 θ) = sin (2 θ) sin (2 θ)

= sin (2 θ) sin (2 θ) + sin (2 θ) (1 + cos (2 θ))

Factor out common term

sin (2 θ)

= sin (2 θ) (sin (2 θ) + 1 + cos (2 θ))

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

Cancel the common factor:

sin (2 θ) + 1 + cos (2 θ)

= \frac{sin ( 2 θ )}{1 + cos ( 2 θ )}

= \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

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Frequently Asked Questions (FAQ)

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