Solution 4.3:4d
From Förberedande kurs i matematik 1
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- | With the formula for double angles and the Pythagorean identity | + | With the formula for double angles and the Pythagorean identity <math>\cos^2\!v + \sin^2\!v = 1</math>, we can express <math>\cos 2v</math> in terms of <math>\cos v</math>, |
- | <math>\cos ^ | + | |
- | <math>\ | + | |
- | in terms of | + | |
- | <math>\ | + | |
- | + | {{Displayed math||<math>\begin{align} | |
- | <math>\begin{align} | + | \cos 2v &= \cos^2\!v - \sin^2\!v\\[5pt] |
- | + | &= \cos^2\!v - (1-\cos^2\!v)\\[5pt] | |
- | & =2\cos ^ | + | &= 2\cos^2\!v-1\\[5pt] |
- | \end{align}</math> | + | &= 2b^2-1\,\textrm{.} |
+ | \end{align}</math>}} |
Current revision
With the formula for double angles and the Pythagorean identity \displaystyle \cos^2\!v + \sin^2\!v = 1, we can express \displaystyle \cos 2v in terms of \displaystyle \cos v,
\displaystyle \begin{align}
\cos 2v &= \cos^2\!v - \sin^2\!v\\[5pt] &= \cos^2\!v - (1-\cos^2\!v)\\[5pt] &= 2\cos^2\!v-1\\[5pt] &= 2b^2-1\,\textrm{.} \end{align} |