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

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

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Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{1 - tan ( x )}{1 + tan ( x )}

Express with sin, cos

\frac{1 - tan ( x )}{1 + tan ( x )}

Use the basic trigonometric identity:

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

= \frac{1 - \frac{s i n ( x )}{c o s ( x )}}{1 + \frac{s i n ( x )}{c o s ( x )}}

Simplify

\frac{1 - \frac{s i n ( x )}{c o s ( x )}}{1 + \frac{s i n ( x )}{c o s ( x )}} : \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )}

\frac{1 - \frac{s i n ( x )}{c o s ( x )}}{1 + \frac{s i n ( x )}{c o s ( x )}}

Join

1 + \frac{sin ( x )}{cos ( x )} : \frac{cos ( x ) + sin ( x )}{cos ( x )}

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

Convert element to fraction:

1 = \frac{1 cos ( x )}{cos ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

= \frac{1 - \frac{s i n ( x )}{c o s ( x )}}{\frac{c o s ( x ) + s i n ( x )}{c o s ( x )}}

Join

1 - \frac{sin ( x )}{cos ( x )} : \frac{cos ( x ) - sin ( x )}{cos ( x )}

1 - \frac{sin ( x )}{cos ( x )}

Convert element to fraction:

1 = \frac{1 cos ( x )}{cos ( x )}

= \frac{1 \cdot cos ( x )}{cos ( x )} - \frac{sin ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot cos ( x ) - sin ( x )}{cos ( x )}

Multiply:

1 \cdot cos (x) = cos (x)

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

= \frac{\frac{c o s ( x ) - s i n ( x )}{c o s ( x )}}{\frac{c o s ( x ) + s i n ( x )}{c o s ( x )}}

Divide fractions:

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

= \frac{( cos ( x ) - sin ( x ) ) cos ( x )}{cos ( x ) ( cos ( x ) + sin ( x ) )}

Cancel the common factor:

cos (x)

= \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )}

= \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )}

= \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )}

Multiply by

\frac{cos ( 2 x ) ( cos ( x ) - sin ( x ) )}{cos ( 2 x ) ( cos ( x ) - sin ( x ) )}

= \frac{( cos ( x ) - sin ( x ) ) ( cos ( x ) - sin ( x ) ) cos ( 2 x )}{( cos ( x ) + sin ( x ) ) ( cos ( x ) - sin ( x ) ) cos ( 2 x )}

Expand

(cos (x) - sin (x)) (cos (x) - sin (x)) cos (2 x) : cos^{2} (x) cos (2 x) - 2 cos (2 x) cos (x) sin (x) + sin^{2} (x) cos (2 x)

(cos (x) - sin (x)) (cos (x) - sin (x)) cos (2 x)

Apply exponent rule:

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

(cos (x) - sin (x)) (cos (x) - sin (x)) = (cos (x) - sin (x))^{1 + 1}

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

Add the numbers:

1 + 1 = 2

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

(cos (x) - sin (x))^{2} = cos^{2} (x) - 2 cos (x) sin (x) + sin^{2} (x)

(cos (x) - sin (x))^{2}

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = cos (x), b = sin (x)

= cos^{2} (x) - 2 cos (x) sin (x) + sin^{2} (x)

= cos (2 x) (cos^{2} (x) + sin^{2} (x) - 2 cos (x) sin (x))

= cos (2 x) (cos^{2} (x) - 2 cos (x) sin (x) + sin^{2} (x))

Distribute parentheses

= cos (2 x) cos^{2} (x) + cos (2 x) (- 2 cos (x) sin (x)) + cos (2 x) sin^{2} (x)

Apply minus-plus rules

+ (- a) = - a

= cos^{2} (x) cos (2 x) - 2 cos (2 x) cos (x) sin (x) + sin^{2} (x) cos (2 x)

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

Rewrite using trig identities

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

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

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

Simplify

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

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

Factor

cos (2 x) cos^{2} (x) + cos (2 x) (1 - cos^{2} (x)) - 2 cos (2 x) cos (x) sin (x) : cos (2 x) (1 - 2 cos (x) sin (x))

cos (2 x) cos^{2} (x) + cos (2 x) (1 - cos^{2} (x)) - 2 cos (2 x) cos (x) sin (x)

Factor out common term

cos (2 x)

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

Refine

= cos (2 x) (- 2 cos (x) sin (x) + 1)

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

Cancel the common factor:

cos (2 x)

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

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

Use the Double Angle identity:

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

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

Expand

(cos (x) + sin (x)) (cos (x) - sin (x)) : cos^{2} (x) - sin^{2} (x)

(cos (x) + sin (x)) (cos (x) - sin (x))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = cos (x), b = sin (x)

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

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

Use the Double Angle identity:

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

= \frac{1 - sin ( 2 x )}{cos ( 2 x )}

= \frac{1 - sin ( 2 x )}{cos ( 2 x )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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