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evaluate cosec(10)+cosec(50)-cosec(70)

Guest Feb 27, 2017

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 #4
avatar+18624 
+10

evaluate cosec(10)+cosec(50)-cosec(70)

 

Formula:

\(\begin{array}{|rcll|} \hline \sin(10^{\circ}) = \cos(90^{\circ}-10^{\circ}) = \cos(80^{\circ}) \\ \sin(50^{\circ}) = \cos(90^{\circ}-50^{\circ}) = \cos(40^{\circ}) \\ \sin(70^{\circ}) = \cos(90^{\circ}-70^{\circ}) = \cos(20^{\circ}) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && csc(10^{\circ})+csc(50^{\circ})-csc(70^{\circ}) \\ &=& \frac{1}{\sin(10^{\circ})} + \frac{1}{\sin(50^{\circ})} - \frac{1}{\sin(70^{\circ})} \\ &=& \frac{1}{\cos(80^{\circ})} + \frac{1}{\cos(40^{\circ})} - \frac{1}{\cos(20^{\circ})} \\ &=& \frac{\cos(20^{\circ})\cos(40^{\circ})+\cos(20^{\circ})\cos(80^{\circ})-\cos(40^{\circ})\cos(80^{\circ})} {\cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} \\\\ && \cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ}) \\ &&= \frac{ 2\cdot\sin(20^{\circ}) \cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} {2\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(40^{\circ})\cos(40^{\circ})\cos(80^{\circ})} {2\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(80^{\circ})\cos(80^{\circ})} {4\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(160^{\circ}) } {8\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(20^{\circ}) } {8\cdot\sin(20^{\circ}) } \\ &&= \frac{ 1 } {8} \\\\ &=& 8\cdot [~ \cos(20^{\circ})\cos(40^{\circ})+\cos(20^{\circ})\cos(80^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \\ &=& 8\cdot \{~ \cos(20^{\circ})[\cos(40^{\circ})+\cos(80^{\circ})]-\cos(40^{\circ})\cos(80^{\circ}) ~\} \\\\ && \cos(40^{\circ}) = \cos(60^{\circ}-20^{\circ}) = \cos(60^{\circ})\cos(20^{\circ})+\sin(60^{\circ})\sin(20^{\circ}) \\ && \cos(80^{\circ}) = \cos(60^{\circ}+20^{\circ}) = \cos(60^{\circ})\cos(20^{\circ})-\sin(60^{\circ})\sin(20^{\circ}) \\ && \cos(40^{\circ})+\cos(80^{\circ}) = 2 \cos(60^{\circ})\cos(20^{\circ}) \\\\ &=& 8\cdot [~ \cos(20^{\circ})2 \cos(60^{\circ})\cos(20^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \quad | \quad \cos(60^{\circ}) = \frac12\\ &=& 8\cdot [~ \cos(20^{\circ})\cos(20^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \\\\ && \cos(40^{\circ}) = \cos(80^{\circ}-40^{\circ}) = \cos(80^{\circ})\cos(40^{\circ})+\sin(80^{\circ})\sin(40^{\circ}) \\ && -\cos(60^{\circ})=\cos(120^{\circ}) = \cos(80^{\circ}+40^{\circ}) = \cos(80^{\circ})\cos(40^{\circ})-\sin(80^{\circ})\sin(40^{\circ}) \\ && \cos(40^{\circ})-\cos(60^{\circ})= 2\cos(80^{\circ})\cos(40^{\circ}) \\\\ &=& 8\cdot \{~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot [\cos(40^{\circ})-\cos(60^{\circ})] ~\} \\ &=& 8\cdot \{~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot \cos(40^{\circ})+\frac12 \cos(60^{\circ}) ~\} \quad | \quad \cos(60^{\circ}) = \frac12 \\ &=& 8\cdot [~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot \cos(40^{\circ})+\frac14 ~] \\\\ && 1 = \cos(20^{\circ}-20^{\circ}) = \cos(20^{\circ})\cos(20^{\circ})+\sin(20^{\circ})\sin(20^{\circ}) \\ && \cos(40^{\circ}) = \cos(20^{\circ}+20^{\circ}) = \cos(20^{\circ})\cos(20^{\circ})-\sin(20^{\circ})\sin(20^{\circ}) \\ && 1+\cos(40^{\circ}) = 2 \cos(20^{\circ})\cos(20^{\circ}) \\\\ &=& 8\cdot \{~ \frac12[1+\cos(40^{\circ})] -\frac12 \cdot \cos(40^{\circ})+\frac14 ~ \} \\ &=& 8\cdot [~ \frac12 + \frac12 \cdot \cos(40^{\circ}) -\frac12 \cdot \cos(40^{\circ})+\frac14 ~] \\ &=& 8\cdot (~ \frac12 +\frac14 ~) \\ &=& 8\cdot (~ \frac34 ~) \\ &=& \mathbf{6} \\ \hline \end{array} \)

 

laugh

heureka  Feb 27, 2017
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 #1
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evaluate cosec(10)+cosec(50)-cosec(70)

=6

Guest Feb 27, 2017
 #2
avatar+90590 
0

cosec(10)+cosec(50)-cosec(70)=6

 

This is true but I would like to see how a mathematician can prove it as at present it has me baffled.

Melody  Feb 27, 2017
 #3
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+5

PROOF:
cosec 10 + cosec 50 - cosec 70
sec 80 + sec 40 - sec 20
(1/cos 80) + (1/cos 40) - (1/cos 20)
(cos 40 + cos 80) / (cos 80 cos 40) - (1/cos 20)
Apply cos A + cos B = 2 cos ((A + B)/2) cos ((A - B)/2)
(2 cos 60 cos 20) / (cos 40 cos 80) - (1/cos 20)
(cos 20) / (cos 40 cos 80) - (1/cos 20)
Take LCM and add both
(cos²20 - cos 40 cos 80) / (cos 20 cos 40 cos 80)

I am gonna solve the denominator seperately,
cos 20 cos 40 cos 80
Multiply and divide by 2 sin 20
(2 sin 20 cos 20 cos 40 cos 80) / (2 sin 20)
2 sin A cos A = sin 2A
(sin 40 cos 40 cos 80) / (2 sin 20)
Multiply and divide by 2
(2 sin 40 cos 40 cos 80) / (4 sin 20)
(sin 80 cos 80) / (4 sin 20)
Again 2,
sin 160 / (8 sin 20)
sin (180 - 20) = sin 20
sin 160 = sin 20
= 1/8
So, denominator is equal to 1/8

Our expression becomes,
8 (cos²20 - cos 40 cos 80)
Take one 2 inside and use 2 cos A cos B = cos ((A + B)/2) + cos ((A - B)/2)
4 (2cos²20 - (cos 120 + cos 40))
2cos²A = 1 + cos 2A
4 (1 + cos 40 - cos 120 - cos 40)
4 (1 - cos 120)
4 (1 - (-1/2))
4 (3/2) = 6

Guest Feb 27, 2017
 #4
avatar+18624 
+10
Best Answer

evaluate cosec(10)+cosec(50)-cosec(70)

 

Formula:

\(\begin{array}{|rcll|} \hline \sin(10^{\circ}) = \cos(90^{\circ}-10^{\circ}) = \cos(80^{\circ}) \\ \sin(50^{\circ}) = \cos(90^{\circ}-50^{\circ}) = \cos(40^{\circ}) \\ \sin(70^{\circ}) = \cos(90^{\circ}-70^{\circ}) = \cos(20^{\circ}) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && csc(10^{\circ})+csc(50^{\circ})-csc(70^{\circ}) \\ &=& \frac{1}{\sin(10^{\circ})} + \frac{1}{\sin(50^{\circ})} - \frac{1}{\sin(70^{\circ})} \\ &=& \frac{1}{\cos(80^{\circ})} + \frac{1}{\cos(40^{\circ})} - \frac{1}{\cos(20^{\circ})} \\ &=& \frac{\cos(20^{\circ})\cos(40^{\circ})+\cos(20^{\circ})\cos(80^{\circ})-\cos(40^{\circ})\cos(80^{\circ})} {\cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} \\\\ && \cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ}) \\ &&= \frac{ 2\cdot\sin(20^{\circ}) \cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} {2\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(40^{\circ})\cos(40^{\circ})\cos(80^{\circ})} {2\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(80^{\circ})\cos(80^{\circ})} {4\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(160^{\circ}) } {8\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(20^{\circ}) } {8\cdot\sin(20^{\circ}) } \\ &&= \frac{ 1 } {8} \\\\ &=& 8\cdot [~ \cos(20^{\circ})\cos(40^{\circ})+\cos(20^{\circ})\cos(80^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \\ &=& 8\cdot \{~ \cos(20^{\circ})[\cos(40^{\circ})+\cos(80^{\circ})]-\cos(40^{\circ})\cos(80^{\circ}) ~\} \\\\ && \cos(40^{\circ}) = \cos(60^{\circ}-20^{\circ}) = \cos(60^{\circ})\cos(20^{\circ})+\sin(60^{\circ})\sin(20^{\circ}) \\ && \cos(80^{\circ}) = \cos(60^{\circ}+20^{\circ}) = \cos(60^{\circ})\cos(20^{\circ})-\sin(60^{\circ})\sin(20^{\circ}) \\ && \cos(40^{\circ})+\cos(80^{\circ}) = 2 \cos(60^{\circ})\cos(20^{\circ}) \\\\ &=& 8\cdot [~ \cos(20^{\circ})2 \cos(60^{\circ})\cos(20^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \quad | \quad \cos(60^{\circ}) = \frac12\\ &=& 8\cdot [~ \cos(20^{\circ})\cos(20^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \\\\ && \cos(40^{\circ}) = \cos(80^{\circ}-40^{\circ}) = \cos(80^{\circ})\cos(40^{\circ})+\sin(80^{\circ})\sin(40^{\circ}) \\ && -\cos(60^{\circ})=\cos(120^{\circ}) = \cos(80^{\circ}+40^{\circ}) = \cos(80^{\circ})\cos(40^{\circ})-\sin(80^{\circ})\sin(40^{\circ}) \\ && \cos(40^{\circ})-\cos(60^{\circ})= 2\cos(80^{\circ})\cos(40^{\circ}) \\\\ &=& 8\cdot \{~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot [\cos(40^{\circ})-\cos(60^{\circ})] ~\} \\ &=& 8\cdot \{~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot \cos(40^{\circ})+\frac12 \cos(60^{\circ}) ~\} \quad | \quad \cos(60^{\circ}) = \frac12 \\ &=& 8\cdot [~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot \cos(40^{\circ})+\frac14 ~] \\\\ && 1 = \cos(20^{\circ}-20^{\circ}) = \cos(20^{\circ})\cos(20^{\circ})+\sin(20^{\circ})\sin(20^{\circ}) \\ && \cos(40^{\circ}) = \cos(20^{\circ}+20^{\circ}) = \cos(20^{\circ})\cos(20^{\circ})-\sin(20^{\circ})\sin(20^{\circ}) \\ && 1+\cos(40^{\circ}) = 2 \cos(20^{\circ})\cos(20^{\circ}) \\\\ &=& 8\cdot \{~ \frac12[1+\cos(40^{\circ})] -\frac12 \cdot \cos(40^{\circ})+\frac14 ~ \} \\ &=& 8\cdot [~ \frac12 + \frac12 \cdot \cos(40^{\circ}) -\frac12 \cdot \cos(40^{\circ})+\frac14 ~] \\ &=& 8\cdot (~ \frac12 +\frac14 ~) \\ &=& 8\cdot (~ \frac34 ~) \\ &=& \mathbf{6} \\ \hline \end{array} \)

 

laugh

heureka  Feb 27, 2017

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