Is cos(20pi-x)+sin((83pi)/2+x)=0 ?

The answer to whether cos(20pi-x)+sin((83pi)/2+x)=0 is True

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prove cos(20pi-x)+sin((83pi)/2+x)=0

Solution

prove

cos (20 π - x) + sin (\frac{83 π}{2} + x) = 0

Solution

True

Solution steps

cos (20 π - x) + sin (\frac{83 π}{2} + x) = 0

Manipulating left side

cos (20 π - x) + sin (\frac{83 π}{2} + x)

Rewrite using trig identities

sin (\frac{83 π}{2} + x)

Use the Angle Sum identity:

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

= sin (\frac{83 π}{2}) cos (x) + cos (\frac{83 π}{2}) sin (x)

Simplify

sin (\frac{83 π}{2}) cos (x) + cos (\frac{83 π}{2}) sin (x) : - cos (x)

sin (\frac{83 π}{2}) cos (x) + cos (\frac{83 π}{2}) sin (x)

sin (\frac{83 π}{2}) cos (x) = - cos (x)

sin (\frac{83 π}{2}) cos (x)

sin (\frac{83 π}{2}) = - 1

sin (\frac{83 π}{2})

Rewrite using trig identities:

2 sin (\frac{3 π}{4}) cos (\frac{3 π}{4})

sin (\frac{83 π}{2})

Write

sin (\frac{83 π}{2})

sin (2 \cdot \frac{83 π}{4})

= sin (2 \cdot \frac{83 π}{4})

Use the Double Angle identity:

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

= 2 sin (\frac{83 π}{4}) cos (\frac{83 π}{4})

sin (\frac{83 π}{4}) = sin (\frac{3 π}{4})

sin (\frac{83 π}{4})

Rewrite

\frac{83 π}{4}

2 π \cdot 10 + \frac{3 π}{4}

= sin (2 π 10 + \frac{3 π}{4})

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 10 + \frac{3 π}{4}) = sin (\frac{3 π}{4})

= sin (\frac{3 π}{4})

= 2 sin (\frac{3 π}{4}) cos (\frac{83 π}{4})

cos (\frac{83 π}{4}) = cos (\frac{3 π}{4})

cos (\frac{83 π}{4})

Rewrite

\frac{83 π}{4}

2 π \cdot 10 + \frac{3 π}{4}

= cos (2 π 10 + \frac{3 π}{4})

Apply the periodicity of

cos

cos (x + 2 π \cdot k) = cos (x)

cos (2 π \cdot 10 + \frac{3 π}{4}) = cos (\frac{3 π}{4})

= cos (\frac{3 π}{4})

= 2 sin (\frac{3 π}{4}) cos (\frac{3 π}{4})

= 2 sin (\frac{3 π}{4}) cos (\frac{3 π}{4})

Use the following trivial identity:

sin (\frac{3 π}{4}) = \frac{2}{2}

sin (\frac{3 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}

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 π}{4})

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}

= 2 \cdot \frac{2}{2} (- \frac{2}{2})

Simplify

2 \cdot \frac{2}{2} (- \frac{2}{2}) : - 1

2 \cdot \frac{2}{2} (- \frac{2}{2})

Remove parentheses:

(- a) = - a

= - 2 \cdot \frac{2}{2} \cdot \frac{2}{2}

Multiply fractions:

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

= - \frac{2 2 \cdot 2}{2 \cdot 2}

Cancel the common factor:

2

= - \frac{2 2}{2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

= - 1

= - 1 \cdot cos (x)

Multiply:

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

= - cos (x)

= - cos (x) + cos (\frac{83 π}{2}) sin (x)

cos (\frac{83 π}{2}) sin (x) = 0

cos (\frac{83 π}{2}) sin (x)

cos (\frac{83 π}{2}) = 0

cos (\frac{83 π}{2})

Rewrite using trig identities:

1 - 2 sin^{2} (\frac{3 π}{4})

cos (\frac{83 π}{2})

Write

cos (\frac{83 π}{2})

cos (2 \cdot \frac{83 π}{4})

= cos (2 \cdot \frac{83 π}{4})

Use the Double Angle identity:

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

= 1 - 2 sin^{2} (\frac{83 π}{4})

sin (\frac{83 π}{4}) = sin (\frac{3 π}{4})

sin (\frac{83 π}{4})

Rewrite

\frac{83 π}{4}

2 π \cdot 10 + \frac{3 π}{4}

= sin (2 π 10 + \frac{3 π}{4})

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 10 + \frac{3 π}{4}) = sin (\frac{3 π}{4})

= sin (\frac{3 π}{4})

= 1 - 2 sin^{2} (\frac{3 π}{4})

= 1 - 2 sin^{2} (\frac{3 π}{4})

Use the following trivial identity:

sin (\frac{3 π}{4}) = \frac{2}{2}

sin (\frac{3 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}

= 1 - 2 (\frac{2}{2})^{2}

Simplify

1 - 2 (\frac{2}{2})^{2} : 0

1 - 2 (\frac{2}{2})^{2}

2 (\frac{2}{2})^{2} = 1

2 (\frac{2}{2})^{2}

(\frac{2}{2})^{2} = \frac{1}{2}

(\frac{2}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 2 ) ^{2}}{2 ^{2}}

(2)^{2} : 2

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 2

= \frac{2}{2 ^{2}}

Cancel the common factor:

2

= \frac{1}{2}

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

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 1 - 1

Subtract the numbers:

1 - 1 = 0

= 0

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

= - cos (x) + 0

- cos (x) + 0 = - cos (x)

= - cos (x)

= - cos (x)

= cos (20 π - x) - cos (x)

Rewrite using trig identities

cos (20 π - x)

Use the Angle Difference identity:

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

= cos (20 π) cos (x) + sin (20 π) sin (x)

Simplify

cos (20 π) cos (x) + sin (20 π) sin (x) : cos (x)

cos (20 π) cos (x) + sin (20 π) sin (x)

cos (20 π) cos (x) = cos (x)

cos (20 π) cos (x)

cos (20 π) = 1

cos (20 π)

cos (20 π) = cos (0)

cos (20 π)

Rewrite

20 π

2 π \cdot 10 + 0

= cos (2 π 10 + 0)

Apply the periodicity of

cos

cos (x + 2 π \cdot k) = cos (x)

cos (2 π \cdot 10 + 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 cos (x)

Multiply:

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

= cos (x)

sin (20 π) sin (x) = 0

sin (20 π) sin (x)

sin (20 π) = 0

sin (20 π)

sin (20 π) = sin (0)

sin (20 π)

Rewrite

20 π

2 π \cdot 10 + 0

= sin (2 π 10 + 0)

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 10 + 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 sin (x)

Apply rule

0 \cdot a = 0

= 0

= cos (x) + 0

cos (x) + 0 = cos (x)

= cos (x)

= cos (x)

= cos (x) - cos (x)

Add similar elements:

cos (x) - cos (x) = 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is cos(20pi-x)+sin((83pi)/2+x)=0 ?
The answer to whether cos(20pi-x)+sin((83pi)/2+x)=0 is True