Is cos(-pi/6)=sin(-pi/6+pi/2) ?

The answer to whether cos(-pi/6)=sin(-pi/6+pi/2) is True

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prove cos(-pi/6)=sin(-pi/6+pi/2)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

cos (- \frac{π}{6})

Use the negative angle identity:

cos (- x) = cos (x)

= cos (\frac{π}{6})

Simplify

= \frac{3}{2}

Manipulating right side

sin (- \frac{π}{6} + \frac{π}{2})

Rewrite using trig identities

sin (- \frac{π}{6} + \frac{π}{2})

Use the Angle Difference identity:

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

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

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

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

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

sin (\frac{π}{2}) cos (\frac{π}{6})

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

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}

= 1

= 1 \cdot cos (\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}

= 1 \cdot \frac{3}{2}

Multiply:

1 \cdot \frac{3}{2} = \frac{3}{2}

= \frac{3}{2}

cos (\frac{π}{2}) sin (\frac{π}{6}) = 0

cos (\frac{π}{2}) sin (\frac{π}{6})

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot sin (\frac{π}{6})

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}

= 0 \cdot \frac{1}{2}

Apply rule

0 \cdot a = 0

= 0

= \frac{3}{2} - 0

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

= \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2}

We showed that the two sides could take the same form

\Rightarrow True

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Is cos(-pi/6)=sin(-pi/6+pi/2) ?
The answer to whether cos(-pi/6)=sin(-pi/6+pi/2) is True