Is cos(-θ+pi/2)=sin(θ) ?

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

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prove cos(-θ+pi/2)=sin(θ)

prove

cos (- θ + \frac{π}{2}) = sin (θ)

True

Solution steps

cos (- θ + \frac{π}{2}) = sin (θ)

Manipulating left side

cos (- θ + \frac{π}{2})

Rewrite using trig identities

cos (- θ + \frac{π}{2})

Use the Angle Difference identity:

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

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

Simplify

cos (\frac{π}{2}) cos (θ) + sin (\frac{π}{2}) sin (θ) : sin (θ)

cos (\frac{π}{2}) cos (θ) + sin (\frac{π}{2}) sin (θ)

cos (\frac{π}{2}) cos (θ) = 0

cos (\frac{π}{2}) cos (θ)

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 cos (θ)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (θ) = sin (θ)

sin (\frac{π}{2}) sin (θ)

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 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

\frac{sin ( x )}{1 - cos ( x )} - csc (x) = cot (x)

- 1 - cot^{2} (x) = - csc^{2} (x)

sin (2 x) + cos (x) = 0

cot (x) (tan (x) sin (x) + cos (x)) = csc (x)

1 - sin (2 x) csc (x) = cos (x) cot (x)