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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

cos (\frac{π}{6} + x)

Use the Angle Sum identity:

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

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

Simplify

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

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

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

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

Rewrite using trig identities

sin (\frac{π}{3} + x)

Use the Angle Sum identity:

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

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

Simplify

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

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

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

\frac{3}{2} cos (x) - \frac{3}{2} cos (x) = 0

\frac{3}{2} cos (x) - \frac{3}{2} cos (x)

Factor out common term

cos (x)

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

\frac{3}{2} - \frac{3}{2} = 0

\frac{3}{2} - \frac{3}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{3 - 3}{2}

Factor

3 - 3 : 0

3 - 3

Factor out common term

3

= 3 (1 - 1)

Refine

= 0

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

= 0

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

Add similar elements:

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

\frac{1}{2} sin (x) + \frac{1}{2} sin (x)

Factor out common term

sin (x)

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

\frac{1}{2} + \frac{1}{2} = 1

\frac{1}{2} + \frac{1}{2}

Apply rule

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

= \frac{1 + 1}{2}

Refine

= 1

= sin (x)

= sin (x)

= sin (x)

= sin (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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