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:

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

$\blacksquare$