Is (sin(pi-x))/(sin(pi/2-x))=tan(x) ?

The answer to whether (sin(pi-x))/(sin(pi/2-x))=tan(x) is True

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prove (sin(pi-x))/(sin(pi/2-x))=tan(x)

Solution

prove

\frac{sin ( π - x )}{sin ( \frac{π}{2} - x )} = tan (x)

Solution

True

Solution steps

\frac{sin ( π - x )}{sin ( \frac{π}{2} - x )} = tan (x)

Manipulating left side

\frac{sin ( π - x )}{sin ( \frac{π}{2} - x )}

Rewrite using trig identities

sin (\frac{π}{2} - x)

Use the Angle Difference identity:

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

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

Simplify

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

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

sin (\frac{π}{2}) cos (x) = cos (x)

sin (\frac{π}{2}) cos (x)

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 (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

cos (\frac{π}{2}) sin (x) = 0

cos (\frac{π}{2}) sin (x)

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 (x)

Apply rule

0 \cdot a = 0

= 0

= cos (x) - 0

cos (x) - 0 = cos (x)

= cos (x)

= cos (x)

= \frac{sin ( π - x )}{cos ( x )}

Rewrite using trig identities

sin (π - x)

Use the Angle Difference identity:

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

= sin (π) cos (x) - cos (π) sin (x)

Simplify

sin (π) cos (x) - cos (π) sin (x) : sin (x)

sin (π) cos (x) - cos (π) sin (x)

sin (π) cos (x) = 0

sin (π) cos (x)

Simplify

sin (π) : 0

sin (π)

Use the following trivial identity:

sin (π) = 0

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}

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

cos (π) sin (x) = - sin (x)

cos (π) sin (x)

Simplify

cos (π) : - 1

cos (π)

Use the following trivial identity:

cos (π) = (- 1)

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}

= - 1

= - 1 \cdot sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= - sin (x)

= 0 - (- sin (x))

Refine

= sin (x)

= sin (x)

= \frac{sin ( x )}{cos ( x )}

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

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Is (sin(pi-x))/(sin(pi/2-x))=tan(x) ?
The answer to whether (sin(pi-x))/(sin(pi/2-x))=tan(x) is True