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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Simplify

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

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

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

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

Manipulating right side

sin (x - \frac{π}{6})

Rewrite using trig identities

sin (x - \frac{π}{6})

Use the Angle Difference identity:

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

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

Simplify

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

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

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{3}{2} sin (x) - sin (\frac{π}{6}) cos (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{3}{2} sin (x) - \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)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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