Solution 4.3:8c
From Förberedande kurs i matematik 1
(Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_8c-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_3_8c-2(2).gif </center> {{NAVCONTENT_STOP}}) |
m |
||
(3 intermediate revisions not shown.) | |||
Line 1: | Line 1: | ||
- | {{ | + | One could write <math>\tan\frac{u}{2}</math> as a quotient involving sine and cosine, and then continue with the formula for half-angles, |
- | < | + | |
- | {{ | + | {{Displayed math||<math>\tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots</math>}} |
- | {{ | + | |
- | < | + | but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side. |
- | {{ | + | |
+ | We write <math>u</math> as <math>2\cdot(u/2)</math>and use the formula for double angles (so as to end up with a right-hand side which has <math>u/2</math> as its argument), | ||
+ | |||
+ | {{Displayed math||<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}} | ||
+ | |||
+ | Writing the 1 in the denominator as <math>\cos^2(u/2) + \sin^2(u/2)</math> using the Pythagorean identity, | ||
+ | |||
+ | {{Displayed math||<math>\begin{align} | ||
+ | \frac{2\cos\dfrac{u}{2}\cdot\sin\dfrac{u}{2}}{1+\cos^2\dfrac{u}{2}-\sin^2\dfrac{u}{2}} | ||
+ | &= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{\cos^2\dfrac{u}{2} + \sin^2\dfrac{u}{2} + \cos^2\dfrac{u}{2} - \sin^2\dfrac{u}{2}}\\[8pt] | ||
+ | &= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{2\cos^2\dfrac{u}{2}}\\[5pt] | ||
+ | &= \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}}\\[8pt] | ||
+ | &= \tan\frac{u}{2}\,\textrm{.} | ||
+ | \end{align}</math>}} |
Current revision
One could write \displaystyle \tan\frac{u}{2} as a quotient involving sine and cosine, and then continue with the formula for half-angles,
\displaystyle \tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots |
but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.
We write \displaystyle u as \displaystyle 2\cdot(u/2)and use the formula for double angles (so as to end up with a right-hand side which has \displaystyle u/2 as its argument),
\displaystyle \frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.} |
Writing the 1 in the denominator as \displaystyle \cos^2(u/2) + \sin^2(u/2) using the Pythagorean identity,
\displaystyle \begin{align}
\frac{2\cos\dfrac{u}{2}\cdot\sin\dfrac{u}{2}}{1+\cos^2\dfrac{u}{2}-\sin^2\dfrac{u}{2}} &= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{\cos^2\dfrac{u}{2} + \sin^2\dfrac{u}{2} + \cos^2\dfrac{u}{2} - \sin^2\dfrac{u}{2}}\\[8pt] &= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{2\cos^2\dfrac{u}{2}}\\[5pt] &= \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}}\\[8pt] &= \tan\frac{u}{2}\,\textrm{.} \end{align} |