Power Reduction Formulas/Sine to 4th
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Theorem
- $\sin^4 x = \dfrac {3 - 4 \cos 2 x + \cos 4 x} 8$
where $\sin$ and $\cos$ denote sine and cosine respectively.
Proof
\(\ds \sin^4 x\) | \(=\) | \(\ds \paren {\sin^2 x}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac {1 - \cos 2 x} 2}^2\) | Square of Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 - 2 \cos 2 x + \cos^2 2 x} 4\) | multiplying out | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 - 2 \cos 2 x + \frac {1 + \cos 4 x} 2} 4\) | Square of Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 - 4 \cos 2 x + 1 + \cos 4 x} 8\) | multiplying top and bottom by $2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {3 - 4 \cos 2 x + \cos 4 x} 8\) | rearrangement |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.57$