Is tan(pi/4-θ)=(cos(2θ))/(1+sin(2θ)) ?

The answer to whether tan(pi/4-θ)=(cos(2θ))/(1+sin(2θ)) is True

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Popular >

prove tan(pi/4-θ)=(cos(2θ))/(1+sin(2θ))

Solution

prove

tan (\frac{π}{4} - θ) = \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

Solution

True

Solution steps

tan (\frac{π}{4} - θ) = \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

Manipulating left side

tan (\frac{π}{4} - θ)

Rewrite using trig identities

tan (\frac{π}{4} - θ)

Use the basic trigonometric identity:

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

= \frac{sin ( \frac{π}{4} - θ )}{cos ( \frac{π}{4} - θ )}

Use the Angle Difference identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= \frac{sin ( \frac{π}{4} ) cos ( θ ) - cos ( \frac{π}{4} ) sin ( θ )}{cos ( \frac{π}{4} - θ )}

Use the Angle Difference identity:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= \frac{sin ( \frac{π}{4} ) cos ( θ ) - cos ( \frac{π}{4} ) sin ( θ )}{cos ( \frac{π}{4} ) cos ( θ ) + sin ( \frac{π}{4} ) sin ( θ )}

Simplify

\frac{sin ( \frac{π}{4} ) cos ( θ ) - cos ( \frac{π}{4} ) sin ( θ )}{cos ( \frac{π}{4} ) cos ( θ ) + sin ( \frac{π}{4} ) sin ( θ )} : \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

\frac{sin ( \frac{π}{4} ) cos ( θ ) - cos ( \frac{π}{4} ) sin ( θ )}{cos ( \frac{π}{4} ) cos ( θ ) + sin ( \frac{π}{4} ) sin ( θ )}

sin (\frac{π}{4}) cos (θ) - cos (\frac{π}{4}) sin (θ) = \frac{2}{2} cos (θ) - \frac{2}{2} sin (θ)

sin (\frac{π}{4}) cos (θ) - cos (\frac{π}{4}) sin (θ)

Simplify

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{2}{2} cos (θ) - cos (\frac{π}{4}) sin (θ)

Simplify

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

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

= \frac{\frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}{cos ( \frac{π}{4} ) cos ( θ ) + sin ( \frac{π}{4} ) sin ( θ )}

cos (\frac{π}{4}) cos (θ) + sin (\frac{π}{4}) sin (θ) = \frac{2}{2} cos (θ) + \frac{2}{2} sin (θ)

cos (\frac{π}{4}) cos (θ) + sin (\frac{π}{4}) sin (θ)

Simplify

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2} cos (θ) + sin (\frac{π}{4}) sin (θ)

Simplify

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

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

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

Multiply

\frac{2}{2} cos (θ) : \frac{2 cos ( θ )}{2}

\frac{2}{2} cos (θ)

Multiply fractions:

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

= \frac{2 cos ( θ )}{2}

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

Multiply

\frac{2}{2} sin (θ) : \frac{2 sin ( θ )}{2}

\frac{2}{2} sin (θ)

Multiply fractions:

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

= \frac{2 sin ( θ )}{2}

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

Multiply

\frac{2}{2} cos (θ) : \frac{2 cos ( θ )}{2}

\frac{2}{2} cos (θ)

Multiply fractions:

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

= \frac{2 cos ( θ )}{2}

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

Multiply

\frac{2}{2} sin (θ) : \frac{2 sin ( θ )}{2}

\frac{2}{2} sin (θ)

Multiply fractions:

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

= \frac{2 sin ( θ )}{2}

= \frac{\frac{2 c o s ( θ )}{2} - \frac{2 s i n ( θ )}{2}}{\frac{2 c o s ( θ )}{2} + \frac{2 s i n ( θ )}{2}}

Combine the fractions

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

Apply rule

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

= \frac{2 cos ( θ ) + 2 sin ( θ )}{2}

= \frac{\frac{2 c o s ( θ )}{2} - \frac{2 s i n ( θ )}{2}}{\frac{2 c o s ( θ ) + 2 s i n ( θ )}{2}}

Combine the fractions

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

Apply rule

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

= \frac{2 cos ( θ ) - 2 sin ( θ )}{2}

= \frac{\frac{2 c o s ( θ ) - 2 s i n ( θ )}{2}}{\frac{2 c o s ( θ ) + 2 s i n ( θ )}{2}}

Divide fractions:

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

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

Cancel the common factor:

2

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

Factor out common term

2

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

Factor out common term

2

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

Cancel the common factor:

2

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

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

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

Multiply by

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

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

Expand

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

(cos (θ) - sin (θ)) (cos (θ) - sin (θ)) cos (2 θ)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

= (cos (θ) - sin (θ))^{2} cos (2 θ)

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

(cos (θ) - sin (θ))^{2}

Apply Perfect Square Formula:

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

a = cos (θ), b = sin (θ)

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

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

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

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

Factor out common term

cos (2 θ)

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

Refine

= cos (2 θ) (- 2 cos (θ) sin (θ) + 1)

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

Cancel the common factor:

cos (2 θ)

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

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

Use the Double Angle identity:

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

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

Expand

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

(cos (θ) + sin (θ)) (cos (θ) - sin (θ))

Apply Difference of Two Squares Formula:

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

a = cos (θ), b = sin (θ)

= cos^{2} (θ) - sin^{2} (θ)

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

Use the Double Angle identity:

cos^{2} (θ) - sin^{2} (θ) = cos (2 θ)

= \frac{1 - sin ( 2 θ )}{cos ( 2 θ )}

Multiply by

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (2 θ)

= \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

= \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

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Frequently Asked Questions (FAQ)

Is tan(pi/4-θ)=(cos(2θ))/(1+sin(2θ)) ?
The answer to whether tan(pi/4-θ)=(cos(2θ))/(1+sin(2θ)) is True