Is sin(pi/6+x)=cos(pi/3-x) ?

The answer to whether sin(pi/6+x)=cos(pi/3-x) is True

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Popular Trigonometry >

prove sin(pi/6+x)=cos(pi/3-x)

Solution

prove

sin (\frac{π}{6} + x) = cos (\frac{π}{3} - x)

Solution

True

Solution steps

sin (\frac{π}{6} + x) = cos (\frac{π}{3} - x)

Manipulating left side

sin (\frac{π}{6} + x)

Rewrite using trig identities

sin (\frac{π}{6} + x)

Use the Angle Sum identity:

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

= sin (\frac{π}{6}) cos (x) + cos (\frac{π}{6}) sin (x)

Simplify

sin (\frac{π}{6}) cos (x) + cos (\frac{π}{6}) sin (x) : \frac{1}{2} cos (x) + \frac{3}{2} sin (x)

sin (\frac{π}{6}) cos (x) + cos (\frac{π}{6}) sin (x)

Simplify

sin (\frac{π}{6}) : \frac{1}{2}

sin (\frac{π}{6})

Use the following trivial identity:

sin (\frac{π}{6}) = \frac{1}{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{1}{2}

= \frac{1}{2} cos (x) + cos (\frac{π}{6}) sin (x)

Simplify

cos (\frac{π}{6}) : \frac{3}{2}

cos (\frac{π}{6})

Use the following trivial identity:

cos (\frac{π}{6}) = \frac{3}{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{3}{2}

= \frac{1}{2} cos (x) + \frac{3}{2} sin (x)

= \frac{1}{2} cos (x) + \frac{3}{2} sin (x)

= \frac{1}{2} cos (x) + \frac{3}{2} sin (x)

Manipulating right side

cos (\frac{π}{3} - x)

Rewrite using trig identities

cos (\frac{π}{3} - x)

Use the Angle Difference identity:

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

= cos (\frac{π}{3}) cos (x) + sin (\frac{π}{3}) sin (x)

Simplify

cos (\frac{π}{3}) cos (x) + sin (\frac{π}{3}) sin (x) : \frac{1}{2} cos (x) + \frac{3}{2} sin (x)

cos (\frac{π}{3}) cos (x) + sin (\frac{π}{3}) sin (x)

Simplify

cos (\frac{π}{3}) : \frac{1}{2}

cos (\frac{π}{3})

Use the following trivial identity:

cos (\frac{π}{3}) = \frac{1}{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{1}{2}

= \frac{1}{2} cos (x) + sin (\frac{π}{3}) sin (x)

Simplify

sin (\frac{π}{3}) : \frac{3}{2}

sin (\frac{π}{3})

Use the following trivial identity:

sin (\frac{π}{3}) = \frac{3}{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{3}{2}

= \frac{1}{2} cos (x) + \frac{3}{2} sin (x)

= \frac{1}{2} cos (x) + \frac{3}{2} sin (x)

= \frac{1}{2} cos (x) + \frac{3}{2} sin (x)

We showed that the two sides could take the same form

\Rightarrow True

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Is sin(pi/6+x)=cos(pi/3-x) ?
The answer to whether sin(pi/6+x)=cos(pi/3-x) is True