Sine of 45 Degrees

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Theorem

$\sin 45 \degrees = \sin \dfrac \pi 4 = \dfrac {\sqrt 2} 2$

where $\sin$ denotes the sine function.


Proof 1

Sine45.png

Let $ABCD$ be a square of side $r$.

By definition, each angle of $\triangle ABCD$ is equal to $90 \degrees$.

Let $AC$ be a diagonal of $ABCD$.

As $\triangle ABC$ is a right angled triangle, it follows from Pythagoras's Theorem that $AC = \sqrt 2 A B$.

As $AC$ is a bisector of $\angle DAB$ it follows that $\angle CAB = 45 \degrees$.

So by definition of sine function:

$\map \sin {\angle CAB} = \dfrac r {r \sqrt 2} = \dfrac {\sqrt 2} 2$

$\blacksquare$


Proof 2

\(\ds \sin 45 \degrees\) \(=\) \(\ds \map \sin {30 \degrees + 15 \degrees}\)
\(\ds \) \(=\) \(\ds \sin 30 \degrees \cos 15 \degrees + \cos 30 \degrees \sin 15 \degrees\) Sine of Sum
\(\ds \) \(=\) \(\ds \paren {\frac 1 2} \paren {\frac {\sqrt 6 + \sqrt 2} 4} + \paren {\frac {\sqrt 3} 2} \paren {\dfrac {\sqrt 6 - \sqrt 2} 4}\) Sine of $30 \degrees$, Cosine of $15 \degrees$, Cosine of $30 \degrees$, Sine of $15 \degrees$
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {\sqrt 6 + \sqrt 2 + \sqrt 3 \paren {\sqrt 6 - \sqrt 2} }\)
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {\sqrt 3 \sqrt 2 + \sqrt 2 + \sqrt 3 \sqrt 3 \sqrt 2 - \sqrt 3 \sqrt 2}\)
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {\sqrt 2 + 3 \sqrt 2}\)
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {4 \sqrt 2}\)
\(\ds \) \(=\) \(\ds \frac {\sqrt 2} 2\)

$\blacksquare$


Proof 3

\(\ds \sin 45 \degrees\) \(=\) \(\ds \map \sin {3 \times 15 \degrees}\)
\(\ds \) \(=\) \(\ds 3 \sin 15 \degrees - 4 \sin^3 15 \degrees\) Triple Angle Formula for Sine
\(\ds \) \(=\) \(\ds 3 \paren {\frac {\sqrt 6 - \sqrt 2} 4} - 4 \paren {\frac {\sqrt 6 - \sqrt 2} 4}^3\) Sine of $15 \degrees$
\(\ds \) \(=\) \(\ds \frac {3 \sqrt 6 - 3 \sqrt 2} 4 - \frac {6 \sqrt 6 - 18 \sqrt 2 + 6 \sqrt 6 - 2 \sqrt 2} {16}\)
\(\ds \) \(=\) \(\ds \frac {3 \sqrt 6 - 3 \sqrt 2} 4 - \frac {12 \sqrt 6 - 20\sqrt 2} {16}\)
\(\ds \) \(=\) \(\ds \frac {3 \sqrt 6 - 3 \sqrt 2} 4 - \frac {3 \sqrt 6 - 5 \sqrt 2} 4\)
\(\ds \) \(=\) \(\ds \frac {2 \sqrt 2} 4\)
\(\ds \) \(=\) \(\ds \frac {\sqrt 2} 2\)

$\blacksquare$


Proof 4

\(\ds \sin 45 \degrees\) \(=\) \(\ds \map \sin {60 \degrees - 15 \degrees}\)
\(\ds \) \(=\) \(\ds \sin 60 \degrees \cos 15 \degrees - \cos 60 \degrees \sin 15 \degrees\) Sine of Difference
\(\ds \) \(=\) \(\ds \paren {\frac {\sqrt 3} 2} \paren {\frac {\sqrt 6 + \sqrt 2} 4} - \paren {\frac 1 2} \paren {\frac {\sqrt 6 - \sqrt 2} 4}\) Sine of $60 \degrees$, Cosine of $15 \degrees$, Cosine of $60 \degrees$, Sine of $15 \degrees$
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {\sqrt 3 \paren {\sqrt 6 + \sqrt 2} - \paren {\sqrt 6 - \sqrt 2} }\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {3 \sqrt 2 + \sqrt 6 - \sqrt 6 + \sqrt 2}\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {3 \sqrt 2 + \sqrt 2}\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {4 \sqrt 2}\) simplifying
\(\ds \) \(=\) \(\ds \frac {\sqrt 2} 2\)

$\blacksquare$


Proof 5

\(\ds \sin 90 \degrees\) \(=\) \(\ds 1\) Sine of Right Angle
\(\ds \leadsto \ \ \) \(\ds \map \sin {2 \times 45 \degrees}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds 2 \sin 45 \degrees \cos 45 \degrees\) \(=\) \(\ds 1\) Double Angle Formula for Sine
\(\ds \leadsto \ \ \) \(\ds 2 \sin 45 \degrees \map \sin {90 \degrees - 45 \degrees}\) \(=\) \(\ds 1\) Sine of Complement equals Cosine
\(\ds \leadsto \ \ \) \(\ds 2 \sin 45 \degrees \sin 45 \degrees\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds 2 \sin^2 45 \degrees\) \(=\) \(\ds 1\) by $(1)$
\(\ds \leadsto \ \ \) \(\ds \sin 45 \degrees\) \(=\) \(\ds \pm \frac 1 {\sqrt 2}\)


The negative solution is rejected because $45 \degrees$ is an acute angle and Sine of Acute Angle is Positive.


Therefore:

$\sin 45 \degrees = \dfrac 1 {\sqrt 2} = \dfrac {\sqrt 2} 2$

$\blacksquare$


Also presented as

Some sources present this result as:

$\sin 45 \degrees = \dfrac 1 {\sqrt 2}$


Sources

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