Keep in mind that pi/4 and pi/6 are two familiar values on the unit circle. We know the sine and cosine of them, so all we have to do is convert everything to sine and cosine. cot(x) = cos(x)/sin(x), and csc(x) = 1/sin(x)
[1/cot(pi/4)] - [2/csc(pi/6)]
[1/(cos(pi/4)/sin(pi/4)] - [2/(1/sin(pi/6))]
Now, we have complex fractions, which we can make into simple fractions. I won't show the details, but the above should simplify into
sin(pi/4)/cos(pi/4) - 2sin(pi/6)
Now, we solve as normal.
[sqrt(2)/2] / [sqrt(2)/2] - 2(1/2)
1 - 1 = 0
2006-12-21 19:56:01
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answer #1
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answered by Puggy 7
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Well you should know that 1/cot a = tan a & 1/sin a = cosec a
then the question can be put in the form,
[1/cot (pi/4)] -2[1/cosec (pi/6)]
= tan (pi/4) -2.sin (pi/6)
= 1 -2(1/2)
= 0
tan (pi/4) = 1 and sin (pi/6) = 1/2 , can be taken as standard values and verified by constructing right angled triangles.
hope this helps!
2006-12-21 20:19:34
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answer #2
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answered by yasiru89 6
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Csc Pi Over 6
2016-12-10 14:33:17
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answer #3
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answered by ? 4
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what is confusing you?
your exp= tan pi/4 - 2*sin pi/6
= 1- 2(.5) = 0
2006-12-21 19:55:05
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answer #4
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answered by Voldemort 1
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{1/[cot(π/4)]} - {2/[csc(π/6)]}
= tan(π/4) - 2sin(π/6)
= 1 - 2(1/2) = 1 - 1 = 0
A little more familiarity with your trig functions will help you a lot. These are very common angles and the student is expected to know, or be able to quickly calculate, the exact values of trig functions of the angles 0, π/6, π/4, π/3, π/2, π, and multiples thereof.
Also, if you are unsure of the values of tan, cot, csc, and sec, convert them to sin and cos first. Then calculate.
2006-12-21 20:41:17
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answer #5
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answered by Northstar 7
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[1/(cos(pi/4)/sin(pi/4))]-[2/(1/sin(pi/6))]=
[sin(pi/4)/cos(pi/4)]-[(2sin(pi/6)]=
tan(pi/4)-2sin(pi/6)
use the trig exact value table remember tan 45(Pi/4)=1 and sin 30(Pi/6)=sqrt(3)/2
=1-2(sqrt(3)/2)
=1-sqrt(3) or approximately equals
= - 0.732050808 to 9 decimal places
2006-12-21 19:55:02
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answer #6
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answered by Zidane 3
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the answer is 0
2006-12-21 19:53:34
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answer #7
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answered by wtfitsnguyen 2
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