Is 2sin(9pi-x)=2sin(x) ?

The answer to whether 2sin(9pi-x)=2sin(x) is True

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prove 2sin(9pi-x)=2sin(x)

prove

2 sin (9 π - x) = 2 sin (x)

True

Solution steps

2 sin (9 π - x) = 2 sin (x)

Manipulating left side

2 sin (9 π - x)

Rewrite using trig identities

sin (9 π - x)

Use the Angle Difference identity:

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

= sin (9 π) cos (x) - cos (9 π) sin (x)

Simplify

sin (9 π) cos (x) - cos (9 π) sin (x) : sin (x)

sin (9 π) cos (x) - cos (9 π) sin (x)

sin (9 π) cos (x) = 0

sin (9 π) cos (x)

sin (9 π) = 0

sin (9 π)

sin (9 π) = sin (π)

sin (9 π)

Rewrite

9 π

2 π \cdot 4 + π

= sin (2 π 4 + π)

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 4 + π) = sin (π)

= sin (π)

= sin (π)

Use the following trivial identity:

sin (π) = 0

sin (π)

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

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

cos (9 π) sin (x) = - sin (x)

cos (9 π) sin (x)

cos (9 π) = - 1

cos (9 π)

cos (9 π) = cos (π)

cos (9 π)

Rewrite

9 π

2 π \cdot 4 + π

= cos (2 π 4 + π)

Apply the periodicity of

cos

cos (x + 2 π \cdot k) = cos (x)

cos (2 π \cdot 4 + π) = cos (π)

= cos (π)

= cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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

= - 1 \cdot sin (x)

Multiply:

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

= - sin (x)

= 0 - (- sin (x))

Refine

= sin (x)

= sin (x)

= 2 sin (x)

We showed that the two sides could take the same form

\Rightarrow True

1 csc (θ) - sin (θ) = cot (θ) cos (θ)

\frac{sec ( x )}{csc ^{2} ( x )} = sec (x) - cos (x)

\frac{1}{sin ( θ ) + 1} = \frac{1 - sin ( θ )}{cos ^{2} ( θ )}

sin (x) cos (x) tan (x) cot (x) = 1

(1 - cos (x)) = cos (x) - 1