Trig problem cos (arctan 5/12)

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I have to calculate this without using the calculator.
cos (arctan 5/12)

So far i draw a triangle and i have the opposite side to be 5, adjacent to be 12, and hypotenuse to be 13. Please suggest me some hint, thanks.
 
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Integral

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Once you have drawn the triangle it is easy. Mark the angle represented by ArcTan [itex] \frac 5 {12} [/itex]

then use the definition of the cos of that angle to get your answer.
 
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Integral said:
Once you have drawn the triangle it is easy. Mark the angle represented by ArcTan [itex] \frac 5 {12} [/itex]

then use the definition of the cos of that angle to get your answer.
am getting 12/13, am i correct
 

Curious3141

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jacy said:
am getting 12/13, am i correct
Correct. BTW, if you're allowed access to a calculator, you can use that to verify your answer (even if you're not allowed to use the calc to derive the answer).
 
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Curious3141 said:
Correct. BTW, if you're allowed access to a calculator, you can use that to verify your answer (even if you're not allowed to use the calc to derive the answer).

Thanks, should the unit be the length of the sides, since 12/13 is not an angle.
 

Curious3141

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jacy said:
Thanks, should the unit be the length of the sides, since 12/13 is not an angle.
No, trig ratios have no unit. You're dividing a length by a length, so they're dimensionless.
 
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Just for another take on this problem (I find the construction of a rt triangle a bit cumbersome) we could use basic trig. results.

For eg, if you have to do something like cos(arctan(x))
we can proceed by taking arctan(x)=y
so x=tan(y)
[tex]\frac{1}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+tan^2y}} = cos(y)
or y=arccos(\frac{1}{\sqrt{1+x^2}})[/tex]
So cos(arctan(x)) = cos(y) = [itex]\frac{1}{\sqrt{1+x^2}})[/itex]
which gives the answer.This approach works for all such problems.
No messy triangles.

Arun
 
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Thanks everyone for helping me out.
 

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