Solution 4.3:8c

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One could write
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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,
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<math>\tan \frac{u}{2}</math>
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as a quotient involving sine and cosine, and then continue with the formula for half-angles,
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<math>\tan \frac{u}{2}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=...</math>
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{{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.
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.
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We write
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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),
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<math>u</math>
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as
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<math>2\left( \frac{u}{2} \right)</math>
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and use the formula for double angles (so as to end up with a right-hand side which has
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<math>\frac{u}{2}</math>
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as its argument)
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<math>\frac{\sin u}{1+\cos u}=\frac{\sin \left( 2\centerdot \frac{u}{2} \right)}{1+\cos \left( 2\centerdot \frac{u}{2} \right)}=\frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}</math>
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Writing the
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{{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>}}
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<math>\text{1}</math>
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in the denominator as
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<math>\cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2}</math>
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using the Pythagorean identity,
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Writing the 1 in the denominator as <math>\cos^2(u/2) + \sin^2(u/2)</math> using the Pythagorean identity,
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}=\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{\cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2}+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}} \\
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\frac{2\cos\dfrac{u}{2}\cdot\sin\dfrac{u}{2}}{1+\cos^2\dfrac{u}{2}-\sin^2\dfrac{u}{2}}
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& =\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{2\cos ^{2}\frac{u}{2}}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=\tan \frac{u}{2} \\
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&= \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]
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\end{align}</math>
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&= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{2\cos^2\dfrac{u}{2}}\\[5pt]
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&= \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}}\\[8pt]
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&= \tan\frac{u}{2}\,\textrm{.}
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\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}

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