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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

cos (\frac{π}{4} - x)

Use the Angle Difference identity:

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

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

Simplify

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

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

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

cos (\frac{π}{4}) cos (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)

Multiply fractions:

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

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

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

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} 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} + cos (\frac{π}{4} + x)

Rewrite using trig identities

cos (\frac{π}{4} + x)

Use the Angle Sum identity:

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

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

Simplify

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

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

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

cos (\frac{π}{4}) cos (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)

Multiply fractions:

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

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

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

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

Simplify

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

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

2 cos (x) + 2 sin (x) + 2 cos (x) - 2 sin (x) = 22 cos (x)

2 cos (x) + 2 sin (x) + 2 cos (x) - 2 sin (x)

Add similar elements:

2 cos (x) + 2 cos (x) = 22 cos (x)

= 22 cos (x) + 2 sin (x) - 2 sin (x)

Add similar elements:

2 sin (x) - 2 sin (x) = 0

= 22 cos (x)

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

Divide the numbers:

\frac{2}{2} = 1

= 2 cos (x)

= 2 cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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