Is sin((9pi)/4)=sin(1/4 pi+2pi) ?

The answer to whether sin((9pi)/4)=sin(1/4 pi+2pi) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sin((9pi)/4)=sin(1/4 pi+2pi)

Solution

prove

sin (\frac{9 π}{4}) = sin (\frac{1}{4} π + 2 π)

Solution

True

Solution steps

sin (\frac{9 π}{4}) = sin (\frac{1}{4} π + 2 π)

Manipulating left side

sin (\frac{9 π}{4})

Simplify

sin (\frac{9 π}{4}) : \frac{2}{2}

sin (\frac{9 π}{4})

sin (\frac{9 π}{4}) = sin (\frac{π}{4})

sin (\frac{9 π}{4})

Rewrite

\frac{9 π}{4}

2 π + \frac{π}{4}

= sin (2 π + \frac{π}{4})

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

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

= sin (\frac{π}{4})

= sin (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

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}

= \frac{2}{2}

Manipulating right side

sin (\frac{1}{4} π + 2 π)

Rewrite using trig identities

sin (\frac{1}{4} π + 2 π)

Use the Angle Sum identity:

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

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

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

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

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

sin (\frac{1}{4} π) cos (2 π)

Multiply

\frac{1}{4} π : \frac{π}{4}

\frac{1}{4} π

Multiply fractions:

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

= \frac{1 π}{4}

Multiply:

1 π = π

= \frac{π}{4}

= cos (2 π) sin (\frac{π}{4})

cos (2 π) = 1

cos (2 π)

cos (2 π) = cos (0)

cos (2 π)

Rewrite

2 π

2 π + 0

= cos (2 π + 0)

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + 0) = cos (0)

= cos (0)

= cos (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

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}

= 1

= 1

= 1 \cdot sin (\frac{π}{4})

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}

= 1 \cdot \frac{2}{2}

Multiply:

\frac{2}{2} \cdot 1 = \frac{2}{2}

= \frac{2}{2}

cos (\frac{1}{4} π) sin (2 π) = 0

cos (\frac{1}{4} π) sin (2 π)

Multiply

\frac{1}{4} π : \frac{π}{4}

\frac{1}{4} π

Multiply fractions:

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

= \frac{1 π}{4}

Multiply:

1 π = π

= \frac{π}{4}

= sin (2 π) cos (\frac{π}{4})

sin (2 π) = 0

sin (2 π)

sin (2 π) = sin (0)

sin (2 π)

Rewrite

2 π

2 π + 0

= sin (2 π + 0)

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

sin (2 π + 0) = sin (0)

= sin (0)

= sin (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

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}

= 0

= 0

= 0 \cdot cos (\frac{π}{4})

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}

= 0 \cdot \frac{2}{2}

Apply rule

0 \cdot a = 0

= 0

= \frac{2}{2} + 0

\frac{2}{2} + 0 = \frac{2}{2}

= \frac{2}{2}

= \frac{2}{2}

= \frac{2}{2}

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

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

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

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

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

prove 2sin^2(x)-1=1-2cos^2(x)prove

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

prove (tan(x)+cot(x))(sin(x)cos(x))=1 prove

(tan (x) + cot (x)) (sin (x) cos (x)) = 1

prove sin(2x)=sin^2(x)-cos^2(x)prove

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

Frequently Asked Questions (FAQ)

Is sin((9pi)/4)=sin(1/4 pi+2pi) ?
The answer to whether sin((9pi)/4)=sin(1/4 pi+2pi) is True