Is sin(pi*x)+sin(pi*(10-x))=0 ?

The answer to whether sin(pi*x)+sin(pi*(10-x))=0 is True

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prove sin(pix)+sin(pi(10-x))=0

Solution

prove

sin (π \cdot x) + sin (π \cdot (10 - x)) = 0

Solution

True

Solution steps

sin (π x) + sin (π (10 - x)) = 0

Manipulating left side

sin (π x) + sin (π (10 - x))

Rewrite as

= sin (π (10 - x)) + sin (π x)

Rewrite using trig identities

sin (π (10 - x)) + sin (π x)

Use the Sum to Product identity:

sin (s) + sin (t) = 2 sin (\frac{s + t}{2}) cos (\frac{s - t}{2})

= 2 sin (\frac{π ( 10 - x ) + π x}{2}) cos (\frac{π ( 10 - x ) - π x}{2})

2 sin (\frac{π ( 10 - x ) + π x}{2}) cos (\frac{π ( 10 - x ) - π x}{2}) = 0

2 sin (\frac{π ( 10 - x ) + π x}{2}) cos (\frac{π ( 10 - x ) - π x}{2})

\frac{π ( 10 - x ) + π x}{2} = 5 π

\frac{π ( 10 - x ) + π x}{2}

Factor

π (10 - x) + π x : 10 π

π (10 - x) + π x

Factor out common term

π

= π (10 - x + x)

Refine

= 10 π

= \frac{10 π}{2}

Divide the numbers:

\frac{10}{2} = 5

= 5 π

= 2 sin (5 π) cos (\frac{- π x + π ( - x + 10 )}{2})

\frac{π ( 10 - x ) - π x}{2} = π (- x + 5)

\frac{π ( 10 - x ) - π x}{2}

Factor

π (10 - x) - π x : π 2 (5 - x)

π (10 - x) - π x

Factor out common term

π

= π (10 - x - x)

Factor

- x - x + 10 : 2 (5 - x)

10 - x - x

Add similar elements:

- x - x = - 2 x

= 10 - 2 x

Rewrite as

= 2 \cdot 5 - 2 x

Factor out common term

2

= 2 (5 - x)

= 2 π (- x + 5)

= \frac{π 2 ( 5 - x )}{2}

Divide the numbers:

\frac{2}{2} = 1

= π (- x + 5)

= 2 sin (5 π) cos (π (- x + 5))

sin (5 π) = 0

sin (5 π)

sin (5 π) = sin (π)

sin (5 π)

Rewrite

5 π

2 π \cdot 2 + π

= sin (2 π 2 + π)

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 2 + 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

= 2 \cdot 0 \cdot cos (π (- x + 5))

Apply rule

0 \cdot a = 0

= 0

= 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is sin(pi*x)+sin(pi*(10-x))=0 ?
The answer to whether sin(pi*x)+sin(pi*(10-x))=0 is True

prove sin(pi*x)+sin(pi*(10-x))=0

Solution

Solution

Popular Examples

Frequently Asked Questions (FAQ)

Is sin(pi*x)+sin(pi*(10-x))=0 ?

prove sin(pix)+sin(pi(10-x))=0

Is sin(pix)+sin(pi(10-x))=0 ?