Is cos((5pi)/4-x)=-(sqrt(2))/2 (cos(x)+sin(x)) ?

The answer to whether cos((5pi)/4-x)=-(sqrt(2))/2 (cos(x)+sin(x)) is True

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prove cos((5pi)/4-x)=-(sqrt(2))/2 (cos(x)+sin(x))

Solution

prove

cos (\frac{5 π}{4} - x) = - \frac{2}{2} (cos (x) + sin (x))

Solution

True

Solution steps

cos (\frac{5 π}{4} - x) = - \frac{2}{2} (cos (x) + sin (x))

Manipulating left side

cos (\frac{5 π}{4} - x)

Rewrite using trig identities

cos (\frac{5 π}{4} - x)

Use the Angle Difference identity:

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

= cos (\frac{5 π}{4}) cos (x) + sin (\frac{5 π}{4}) sin (x)

Simplify

cos (\frac{5 π}{4}) cos (x) + sin (\frac{5 π}{4}) sin (x) : \frac{- 2 cos ( x ) - 2 sin ( x )}{2}

cos (\frac{5 π}{4}) cos (x) + sin (\frac{5 π}{4}) sin (x)

cos (\frac{5 π}{4}) cos (x) = - \frac{2 cos ( x )}{2}

cos (\frac{5 π}{4}) cos (x)

cos (\frac{5 π}{4}) = - \frac{2}{2}

cos (\frac{5 π}{4})

Rewrite using trig identities:

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

cos (\frac{5 π}{4})

Write

cos (\frac{5 π}{4})

cos (π + \frac{π}{4})

= cos (π + \frac{π}{4})

Use the Angle Sum identity:

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

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

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

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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)

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{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}

Use the following trivial identity:

sin (π) = 0

sin (π)

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

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}

= (- 1) \frac{2}{2} - 0 \cdot \frac{2}{2}

Simplify

= - \frac{2}{2}

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

Multiply fractions:

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

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

= - \frac{2 cos ( x )}{2} + sin (\frac{5 π}{4}) sin (x)

sin (\frac{5 π}{4}) sin (x) = - \frac{2 sin ( x )}{2}

sin (\frac{5 π}{4}) sin (x)

sin (\frac{5 π}{4}) = - \frac{2}{2}

sin (\frac{5 π}{4})

Rewrite using trig identities:

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

sin (\frac{5 π}{4})

Write

sin (\frac{5 π}{4})

sin (π + \frac{π}{4})

= sin (π + \frac{π}{4})

Use the Angle Sum identity:

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

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

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

Use the following trivial identity:

sin (π) = 0

sin (π)

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

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{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}

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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)

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}

= 0 \cdot \frac{2}{2} + (- 1) \frac{2}{2}

Simplify

= - \frac{2}{2}

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

Multiply fractions:

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

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

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

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

Manipulating right side

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

Simplify

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is cos((5pi)/4-x)=-(sqrt(2))/2 (cos(x)+sin(x)) ?
The answer to whether cos((5pi)/4-x)=-(sqrt(2))/2 (cos(x)+sin(x)) is True