Reciprocal of One Plus Cosine/Proof 1
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Theorem
- $\dfrac 1 {1 + \cos x} = \dfrac 1 2 \sec^2 \dfrac x 2$
Proof
\(\ds 1 + \cos x\) | \(=\) | \(\ds \cos 0 + \cos x\) | Cosine of Zero is One | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \map \cos {\dfrac {0 + x} 2} \map \cos {\dfrac {0 - x} 2}\) | Cosine plus Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \map \cos {\dfrac x 2} \map \cos {\dfrac {-x} 2}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \map \cos {\dfrac x 2} \map \cos {\dfrac x 2}\) | Cosine Function is Even | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \map {\cos^2} {\frac x 2}\) | simplifying | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 {1 + \cos x}\) | \(=\) | \(\ds \frac 1 2 \map {\sec^2} {\frac x 2}\) | Definition of Secant Function |
$\blacksquare$