Sine of 45 Degrees/Proof 5
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Theorem
- $\sin 45 \degrees = \sin \dfrac \pi 4 = \dfrac {\sqrt 2} 2$
Proof
\(\ds \sin 90 \degrees\) | \(=\) | \(\ds 1\) | 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:
- $\sin 45 \degrees = \dfrac 1 {\sqrt 2} = \dfrac {\sqrt 2} 2$
$\blacksquare$