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

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

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

prove

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

True

Solution steps

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

Manipulating left side

cos (θ + \frac{π}{4})

Rewrite using trig identities

cos (θ + \frac{π}{4})

Use the Angle Sum identity:

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

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

Simplify

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

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

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

cos (θ) 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}

= \frac{2}{2} cos (θ)

Multiply fractions:

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

= \frac{2 cos ( θ )}{2}

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

sin (θ) 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}

= \frac{2}{2} sin (θ)

Multiply fractions:

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

= \frac{2 sin ( θ )}{2}

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

= \frac{2 cos ( θ ) - 2 sin ( θ )}{2}

Manipulating right side

\frac{2}{2} (cos (θ) - sin (θ))

Simplify

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

We showed that the two sides could take the same form

\Rightarrow True

5 cos^{2} (x) + 7 sin^{2} (x) - 5 = 2 sin^{2} (x)

sin (θ) - sin (θ - 2 π) = 0

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

csc (2 x) = \frac{sec ( x ) csc ( x )}{2}

sec (2 x) = \frac{( sec ^{2} ( x ) )}{2 - sec ^{2} ( x )}

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