Using double angle and compound angles formulae prove,
$$ \frac{1-\cos x}{\sin x} = \tan\frac{x}{2} $$
Can someone please help me figure this question, I have no idea how to approach it?
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asked Jul 10, 2013 at 13:36
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2$\begingroup$ Write it as $(1 - \cos(2y))/\sin(2y) = \tan y$. Do you see then how to proceed? $\endgroup$– Daniel FischerJul 10, 2013 at 13:39
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$\begingroup$ I'll try proceeding, thanks for the tip! $\endgroup$– Red Queen10101Jul 10, 2013 at 13:42
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$\begingroup$ Im revising for a test, and im stuck on this question, and I still don't understand many of the hints/solutions people have given me. $\endgroup$– Red Queen10101Jul 10, 2013 at 14:28
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4 Answers
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Use geometry: $AO= 1$
It is strange that no one has mentioned this drawing yet.
P.S. Note that you can easily extract other trigonometric identites involving $\phi/2, 2\phi $ argument from this picture. For example to get $sin(\phi/2)$ use $ECD$ triangle and Pythagorean theorem to calculate $CD/ED = sin (\phi/2)$
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$\begingroup$ I think, this is what Red Queen10101 want $\endgroup$ Jul 10, 2013 at 14:24
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$\begingroup$ we haven't been taught this at school. $\endgroup$ Jul 10, 2013 at 14:26
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6$\begingroup$ School in which pupils did not draw circles and triangles...O tempora, o mores! $\endgroup$– igumnovJul 10, 2013 at 14:32
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$$\dfrac{1-\cos x}{\sin x}=\dfrac{1-(1-2\sin^2\frac x2)}{2\sin\frac x 2\cos\frac x2}=\dfrac{\sin\frac x2}{\cos\frac x2}=\tan\frac x2$$
answered Jul 10, 2013 at 13:45
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8$\begingroup$ When a problem is marked "homework" please don't answer the problem completely. $\endgroup$ Jul 10, 2013 at 13:47
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1$\begingroup$ THANKS! makes self teaching way easier $\endgroup$ Jul 10, 2013 at 14:51
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Hint: It is easier to show:
$$\frac{1-\cos 2y}{\sin 2y} = \tan y$$
using the formulas for $\cos 2y$ and $\sin 2y$.
answered Jul 10, 2013 at 13:40
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$\begingroup$ hi ive used the formulas for cos2y and sin2y. Im still not catching the drift of continuing on. $\endgroup$ Jul 10, 2013 at 13:52
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$$ \cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2} = 1-2\sin^2\frac{x}{2} $$ and $$ \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} $$ so $$ \frac{1-\cos x}{\sin x} = \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \tan\frac{x}{2} $$
answered Jul 10, 2013 at 13:41
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$\begingroup$ (provided $x\neq 0\mod \pi$, of course) $\endgroup$ Jul 10, 2013 at 13:43
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3$\begingroup$ When a problemis marked "homework" please don't answer the problem completely. $\endgroup$ Jul 10, 2013 at 13:44
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$\begingroup$ Hi, im sorry to bug. Im still not understanding what you did on the first row with cosx=..... $\endgroup$ Jul 10, 2013 at 13:58
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$\begingroup$ You have $$\cos(u+v) = \cos u \cos v - \sin u\sin v$$ plug into this $u=v=\frac{x}{2}$. $\endgroup$ Jul 10, 2013 at 14:01