Double Angle Formulas/Cosine/Proof 4

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Theorem

$\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$


Proof

Double angle sin.png

Consider an isosceles triangle $\triangle ABC$ with base $BC$, and head angle $\angle BAC = 2 \alpha$.

Draw an angle bisector to $\angle BAC$ and name it $AH$.

$\angle BAH = \angle CAH = \alpha$

From Angle Bisector and Altitude Coincide iff Triangle is Isosceles:

$AH \perp BC$

From Law of Cosines:

\(\text {(1)}: \quad\) \(\displaystyle CB^2\) \(=\) \(\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha\)


From Pythagoras's Theorem:

\(\displaystyle AC ^ 2\) \(=\) \(\displaystyle CH^2 + AH^2\) in triangle $\triangle AHC$
\(\text {(2.1)}: \quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle CH^2\) \(=\) \(\displaystyle AC^2 - AH^2\)
\(\displaystyle \) \(\) \(\displaystyle \)
\(\displaystyle AB ^ 2\) \(=\) \(\displaystyle BH^2 + AH^2\) in triangle $\triangle AHB$
\(\text {(2.2)}: \quad\) \(\displaystyle \leadsto \ \ \) \(\displaystyle BH^2\) \(=\) \(\displaystyle BC^2 - AH^2\)


By definition of sine:

\(\text {(3.1)}: \quad\) \(\displaystyle CH\) \(=\) \(\displaystyle AC \sin \alpha\)
\(\text {(3.2)}: \quad\) \(\displaystyle BH\) \(=\) \(\displaystyle AB \sin \alpha\)


By definition of cosine:

$AH = AB \cos \alpha = AC \cos \alpha$


So:

\(\text {(4)}: \quad\) \(\displaystyle AH^2\) \(=\) \(\displaystyle AB \cdot AC \cdot \cos^2 \alpha\)
\(\displaystyle \) \(\) \(\displaystyle \)
\(\displaystyle CH^2\) \(=\) \(\displaystyle AC^2 - AH^2\) $(2.1)$
\(\text {(5.1)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle AC^2 - AB \cdot AC \cdot \cos^2 \alpha\) assigning $(4)$
\(\displaystyle \) \(\) \(\displaystyle \)
\(\displaystyle BH^2\) \(=\) \(\displaystyle AB^2 - AH^2\) $(2.2)$
\(\text {(5.2)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle AB^2 - AB \cdot AC \cdot \cos^2 \alpha\) assigning $(4)$


Now:

\(\displaystyle CB^2\) \(=\) \(\displaystyle (CH + BH)^2\)
\(\displaystyle \) \(=\) \(\displaystyle CH^2 + BH^2 + 2 \cdot CH \cdot BH\) Square of Sum
\(\displaystyle \) \(=\) \(\displaystyle AC^2 - AB \cdot AC \cdot \cos^2 \alpha + AB^2 - AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot CH \cdot BH\) assigning $(5.1)$,$(5.2)$
\(\displaystyle \) \(=\) \(\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot CH \cdot BH\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos^2 \alpha + 2 \cdot AB \cdot AC \cdot \sin^2 \alpha\) assigning $(3.1)$,$(3.2)$
\(\displaystyle \) \(=\) \(\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \paren {\cos^2 \alpha - \sin^2 \alpha}\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha\) equating to $(1)$


Hence we get the equation:

\(\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \paren {\cos^2 \alpha - \sin^2 \alpha}\) \(=\) \(\displaystyle AC^2 + AB^2 - 2 \cdot AB \cdot AC \cdot \cos 2 \alpha\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \cos^2 \alpha - \sin^2 \alpha\) \(=\) \(\displaystyle \cos 2 \alpha\) simplifying

$\blacksquare$