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

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

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Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

tan (\frac{π}{4} + x)

Rewrite using trig identities

tan (\frac{π}{4} + x)

Use the basic trigonometric identity:

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

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

Use the Angle Sum identity:

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

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

Use the Angle Sum identity:

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

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

Simplify

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

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

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

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

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 (x) + cos (\frac{π}{4}) sin (x)

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 (x) + \frac{2}{2} sin (x)

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

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

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

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 (x) - sin (\frac{π}{4}) sin (x)

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

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

Multiply

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

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

Multiply fractions:

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

= \frac{2 cos ( x )}{2}

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

Multiply

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

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

Multiply fractions:

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

= \frac{2 sin ( x )}{2}

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

Multiply

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

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

Multiply fractions:

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

= \frac{2 cos ( x )}{2}

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

Multiply

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

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

Multiply fractions:

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

= \frac{2 sin ( x )}{2}

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

Combine the fractions

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

Apply rule

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

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

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

Combine the fractions

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

Apply rule

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

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

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

Divide fractions:

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

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

Cancel the common factor:

2

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

Factor out common term

2

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

Factor out common term

2

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

Cancel the common factor:

2

= \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 )}

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

Apply Perfect Square Formula:

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

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

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

Apply Difference of Two Squares Formula:

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

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

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

Cancel the common factor:

sin (x) + cos (x)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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