Is cos(270-θ)=-sin(θ) ?

The answer to whether cos(270-θ)=-sin(θ) is True

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Popular Trigonometry >

prove cos(270-θ)=-sin(θ)

Solution

prove

cos (27 0^{\circ} - θ) = - sin (θ)

Solution

True

Solution steps

cos (27 0^{\circ} - θ) = - sin (θ)

Manipulating left side

cos (27 0^{\circ} - θ)

Rewrite using trig identities

cos (27 0^{\circ} - θ)

Use the Angle Difference identity:

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

= cos (27 0^{\circ}) cos (θ) + sin (27 0^{\circ}) sin (θ)

Simplify

cos (27 0^{\circ}) cos (θ) + sin (27 0^{\circ}) sin (θ) : - sin (θ)

cos (27 0^{\circ}) cos (θ) + sin (27 0^{\circ}) sin (θ)

cos (27 0^{\circ}) cos (θ) = 0

cos (27 0^{\circ}) cos (θ)

cos (27 0^{\circ}) = 0

cos (27 0^{\circ})

Rewrite using trig identities:

cos (18 0^{\circ}) cos (9 0^{\circ}) - sin (18 0^{\circ}) sin (9 0^{\circ})

cos (27 0^{\circ})

Write

cos (27 0^{\circ})

cos (18 0^{\circ} + 9 0^{\circ})

= cos (18 0^{\circ} + 9 0^{\circ})

Use the Angle Sum identity:

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

= cos (18 0^{\circ}) cos (9 0^{\circ}) - sin (18 0^{\circ}) sin (9 0^{\circ})

= cos (18 0^{\circ}) cos (9 0^{\circ}) - sin (18 0^{\circ}) sin (9 0^{\circ})

Use the following trivial identity:

cos (18 0^{\circ}) = (- 1)

cos (18 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

Use the following trivial identity:

cos (9 0^{\circ}) = 0

cos (9 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

Use the following trivial identity:

sin (18 0^{\circ}) = 0

sin (18 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

Use the following trivial identity:

sin (9 0^{\circ}) = 1

sin (9 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 0 - 0 \cdot 1

Simplify

= 0

= 0 \cdot cos (θ)

Apply rule

0 \cdot a = 0

= 0

sin (27 0^{\circ}) sin (θ) = - sin (θ)

sin (27 0^{\circ}) sin (θ)

sin (27 0^{\circ}) = - 1

sin (27 0^{\circ})

Rewrite using trig identities:

sin (18 0^{\circ}) cos (9 0^{\circ}) + cos (18 0^{\circ}) sin (9 0^{\circ})

sin (27 0^{\circ})

Write

sin (27 0^{\circ})

sin (18 0^{\circ} + 9 0^{\circ})

= sin (18 0^{\circ} + 9 0^{\circ})

Use the Angle Sum identity:

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

= sin (18 0^{\circ}) cos (9 0^{\circ}) + cos (18 0^{\circ}) sin (9 0^{\circ})

= sin (18 0^{\circ}) cos (9 0^{\circ}) + cos (18 0^{\circ}) sin (9 0^{\circ})

Use the following trivial identity:

sin (18 0^{\circ}) = 0

sin (18 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

Use the following trivial identity:

cos (9 0^{\circ}) = 0

cos (9 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

Use the following trivial identity:

cos (18 0^{\circ}) = (- 1)

cos (18 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

Use the following trivial identity:

sin (9 0^{\circ}) = 1

sin (9 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

= 0 \cdot 0 + (- 1) \cdot 1

Simplify

= - 1

= - 1 \cdot sin (θ)

Multiply:

1 \cdot sin (θ) = sin (θ)

= - sin (θ)

= 0 - sin (θ)

0 - sin (θ) = - sin (θ)

= - sin (θ)

= - sin (θ)

= - sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Is cos(270-θ)=-sin(θ) ?
The answer to whether cos(270-θ)=-sin(θ) is True