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Trignometry
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newton
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10 Feb 2010 20:05:40 IST
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Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Trigonometry


 FIND THE VALUE OF        tan(pi/16) + 2 tan(pi/8)  +  4

the ans is cot pi/16

but please don't back explain assuming cot pi/16 to be the value!


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Anshul Rai
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10 Feb 2010 22:10:32 IST
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 It's quite tough to write those trigonometrical equations here, so I'll just copy prewritten tags:                                                                   tan pi/16 = sin (pi/16) / cos (pi/16)

Now, using half angle formulae, we know that cos pi/16 = sqrt(1/2(1+cospi/8)) =sqrt(1/2(1+1/2sqrt(2+sqrt(2))))=1/2sqrt(2+sqrt(2+sqrt(2)))

Similarly,sin pi/16=sqrt(1/2(1-cospi/8)) = sqrt(1/2(1-1/2sqrt(2+sqrt(2)))) = 1/2sqrt(2-sqrt(2+sqrt(2)))

Thus, tan(pi/(16)) = sqrt((2-sqrt(2+sqrt(2)))/(2+sqrt(2+sqrt(2))))= sqrt(4+2sqrt(2))-sqrt(2)-1.(after some simple rationalization.

We know, tan(pi/8)=sqrt(2)-1

Thus, tan(pi/16) + 2 tan(pi/8)  +  4 = +1+sqrt(2)+sqrt(4+2sqrt(2)), which is simply 1/(tan pi/16) = cot pi /16 Hence, the value is found.


 
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newton
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11 Feb 2010 11:56:58 IST
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ya i agree but is there a better solution?
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newton
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18 Feb 2010 15:05:28 IST
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 got a better soln since all the options were

in the form of cot(pi/16)+x where x is a constan

let us assume cot(pi/16)+x to be the sum of given expresion

cotk-tank=2*cot2k (u can prove it easily)

 

 

 

  tan(pi/16) + 2 tan(pi/8)  +  4 tan(pi/4) = cot(pi/16)+x

 

 x+cot[pi/16] - tan[pi/16] = 2cot[2pi/16]+x

 2cot[2pi/16] - 2tan[pi/]8 =4cot[pi4]

4cot[pi4] - 4 tan[pi/4]=0

adding all we get x=0

 

therefore sum is cot[pi/16]

 

 

isn't it simper?

 

 P?::::::::::P?:::::::::::::::::::P?::::::::::::::::::::::::::::::P?::::::::::::
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newton
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20 Feb 2010 23:46:44 IST
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 thanks for the hat!
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