46 lines
1.1 KiB
TeX
46 lines
1.1 KiB
TeX
\begin{deloppgaver}
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\delo
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\begin{align*}
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cos(x) &= -1 \\
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x &= acos(-1) \\
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x &= \pm \pi + 2n\pi
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\end{align*}
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\begin{minipage}{0.42\textwidth}
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\begin{graphbox}
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\input{figures/4a.tex}
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\end{graphbox}
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\end{minipage}
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Ettersom forskjellen mellom $\pi$ og $-\pi$ er $2\pi$ som er et element i $2n\pi$, kan vi slå sammen svarene og si at
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\[ x = \pi + 2n\pi \]
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\delo
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\[cos(2x)=1-2sin^2(x)\]
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Vi substituerer $cos(2x)=cos^2(x)-sin^2(x)$ og $cos^2(x)+sin^2(x)=1$
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\begin{align*}
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cos^2(x)-sin^2(x) &= cos^2(x)+sin^2(x) - 2sin^2(x) \\
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cos^2(x) &= cos^2(x)+2sin^2(x) - 2sin^2(x) \\
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cos^2(x) &= cos^2(x)
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\end{align*}
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\delo
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\[cos(\frac{\pi}{8})\]
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Vi bruker
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\[ sin(2a) = 2 sin(a)cos(a) \]
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Hvor $a = \frac{\pi}{8}$
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\begin{align*}
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sin\left(\frac{\pi}{4}\right) &= 2sin\left(\frac{\pi}{8}\right)cos\left(\frac{\pi}{8}\right) \\[1em]
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cos\left(\frac{\pi}{8}\right) &= \frac{sin\left(\frac{\pi}{4}\right)}{2sin\left(\frac{\pi}{8}\right)} \\[1em]
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&= \frac{\frac{1}{\sqrt{2}}}{2sin\left(\frac{\pi}{8}\right)}
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\end{align*}
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\end{deloppgaver} |