Find the value of $2(\sin2^{\circ}\tan1^{\circ}+\sin4^{\circ}\tan1^{\circ}+\cdots+\sin178^{\circ}\tan 1^{\circ})$
Problem
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Tags: trigonometry, complex numbers
YESMAths
03.06.2014 18:29
SMOJ wrote:
Find the value of
$2(\sin2^{\circ}\tan1^{\circ}+\sin4^{\circ}\tan1^{\circ}+\cdots+\sin178^{\circ}\tan 1^{\circ})$
AnswerANSWER:$\boxed {2\tan1^{\circ}}$
SolutionDenote $2(\sin2^{\circ}\tan1^{\circ}+\sin4^{\circ}\tan1^{\circ}+\cdots+\sin178^{\circ}\tan 1^{\circ})$ as $%Error. "mathmode" is a bad command.
E$ for convenience. So now we have the identity:
$\sin(180-A)=-\sin A$ and we have
$ E = 2{\tan1}^{\circ}({\sin2}^{\circ}+{\sin4}^{\circ}+\cdots-\sin4^{\circ}-\sin2^{\circ})= 2\tan1^{\circ}(\sin90^{\circ})=2\tan1^{\circ}$
rajai7
03.06.2014 19:05
sin(180-A)=-sinA it is wrong......................sin(180-A)=sinA this is correct..............
YESMAths
03.06.2014 19:12
Thanks for the correction, I will try to provide a correct solution.
rajai7
03.06.2014 19:13
You are welcome............@YESMAths.
MSTang
03.06.2014 23:42
First factor that $\tan 1^\circ$ out of there; we'll deal with it later. Now we're investigating the sum $\sin 2^\circ + \sin 4^\circ + \ldots + \sin 178^\circ.$ Using complex numbers ($e^{ix}$) is probably the best way.
byrmc
04.06.2014 00:12
$ 2(\sin2^{\circ}\tan1^{\circ}+\sin4^{\circ}\tan1^{\circ}+\cdots+\sin178^{\circ}\tan 1^{\circ}) $
$ 2\tan1^{\circ}(\sin2^{\circ}+\sin4^{\circ}+\cdots+\sin88^{\circ}+\sin90^{\circ}+\sin88^{\circ}+\cdots+\sin2^{\circ}) $
$ 4\tan1^{\circ}(\sin2^{\circ}+\sin4^{\circ}+\cdots+\sin88^{\circ}+1) $
$ 4\tan1^{\circ}(\sin2^{\circ}+\sin88^{\circ}+\sin4^{\circ}+\sin86^{\circ}\cdots+\sin44^{\circ}+\sin46^{\circ}+1) $
$ 4\tan1^{\circ}(2\sin45^{\circ}\cos45^{\circ}+2\sin45^{\circ}\cos45^{\circ}+\cdots+2\sin45^{\circ}\cos45^{\circ}+1) $
$ 4\tan1^{\circ}(22+1) $
$92\tan1^{\circ} $
byrmc
04.06.2014 00:20
sry my solutions is wrong
delta
04.06.2014 16:14
Rewrite $\tan 1^\circ$ as $\frac {\sin 1^\circ}{\cos 1^\circ}$ . Now factor out $\frac {1}{\cos 1^\circ}$. we're investigating the sum $\sin 1^\circ\sin 2^\circ + \sin1^\circ\sin 4^\circ + \ldots +\sin 1^\circ \sin 178^\circ.$ Using the formula $2\sin x \sin y=\cos (x-y)-\cos(x+y)$ we obtain telescoping sum.
sqing
04.06.2014 16:33
$2(\sin2^{\circ}\tan1^{\circ}+\sin4^{\circ}\tan1^{\circ}+\cdots+\sin178^{\circ}\tan 1^{\circ})=2$.
shmm
04.06.2014 16:47
I think answer $ 2$
sqing
05.06.2014 02:31
shmm wrote: I think answer $ 2$ You are right.
Haroldo60
23.06.2014 23:12
$S= 2(\sin2^{\circ}\tan1^{\circ}+\sin4^{\circ}\tan1^{\circ}+\cdots+\sin178^{\circ}\tan 1^{\circ}) $ $S= 2.\tan1^{\circ}.(\sin2^{\circ}+\sin4^{\circ}+\cdots+\sin178^{\circ}) $ $S= 2.\tan1^{\circ}.S1 $ $S1= 2.(\sin2^{\circ}+\sin4^{\circ}+\cdots+\sin178^{\circ}) $ $\sin1^{\circ}.S1= 2.\sin1^{\circ}.(\sin2^{\circ}+\sin4^{\circ}+\cdots+\sin178^{\circ}) $ $\S1= 2.\frac{\sin89^{\circ}}{\sin1^{\circ}} $ $\S1= 2.\frac{\cos1^{\circ}}{\sin1^{\circ}} $ ${\S1= 2.\cot1^{\circ}} $ ${S= 2.\cot1^{\circ}} \tan1^{\circ} $ $S= 2$
ARCH999
28.06.2014 13:56
We also get this result by using: $sinA + sin(A+B) + sin(A+2B) + ...... + sin(A+(n-1)B) = \frac{sin(A+\frac{n-1}{2}B)*sin\frac{nB}{2}}{sin\frac{B}{2}}$, where $A=B=2$ and $n=89$.