How do you find the 6 trigonometric functions for 16pi/3?

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Answer 1 · 2015-08-12T11:30:22

$sin((16pi)/3)=-sqrt(3)/2$
$cos((16pi)/3)=-1/2$
$tan ((16pi)/3)=sqrt(3)$
$ctg ((16pi)/3)=sqrt(3)/3$
$sec((16pi)/3)=-2$
$csc((16pi)/3)=-(2sqrt(3))/2$

First you have to notice, that the angle is greater than $2pi$ , so you have to reduce it first.

$(16pi)/3=5 1/3 pi=4pi+(4pi)/3$ , so for all functions you can calculate them for $(4pi)/3$ instead of the original value.

Now you can start calculating the values of trig functions:

$sin((4pi)/3)=sin(pi+pi/3)=-sin(pi/3)=-sqrt(3)/2$
$cos((4pi)/3)=cos(pi+pi/3)=-cos(pi/3)=-1/2$

To calculate these values I used 2 facts:

$(4pi)/3$ is in the 3rd quadrant, so tangent and cotangent are positive, other functions are negative
If you have an even multiple of $pi/2$ in the reduction formula, function does not change to a co-function.

Now we can use the trigonometric identities to calculate other 4 functions:

$tan ((4pi)/3)=(sin((4pi)/3))/(cos((4pi)/3))=sqrt(3)$

$ctg ((4pi)/3)=1/tan((4pi)/3)=sqrt(3)/3$

$sec ((4pi)/3)=1/cos ((4pi)/3) = -2$

$csc ((4pi)/3)=1/ sin ((4pi)/3)=-(2sqrt(3))/3$