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
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
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Special angles
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Appendix $12$: Trigonometric formulae: Trigonometric values for some special angles
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $14$: Trigonometric formulae: Trigonometric values for some special angles