How do you give the six trigonometric function values of pi/3?

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How do you give the six trigonometric function values of pi/3?

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Answer 1 · 2017-08-06T19:26:57

$see explanation$ $"see explanation"$

Explanation:

$using the right triangle with angles and sides$ $"using the right triangle with angles and sides"$

$\frac{π}{2}, \frac{π}{6}, \frac{π}{3} \leftarrow angles$ $pi/2,pi/6,pi/3larrcolor(red)" angles"$

$1, \sqrt{3}, 2 \leftarrow sides$ $1,sqrt3,2larrcolor(red)" sides"$

$in relation to cos (\frac{π}{3})$ $"in relation to "cos(pi/3)$

$1 is adjacent, \sqrt{3} is opposite, 2 is hypotenuse$ $1" is adjacent ",sqrt3" is opposite",2" is hypotenuse"$

$∙ x cos (\frac{π}{3}) = \frac{1}{2}$ $•color(white)(x)cos(pi/3)=1/2$

$∙ x sec (\frac{π}{3}) = \frac{1}{cos (\frac{π}{3})} = \frac{1}{\frac{1}{2}} = 2$ $•color(white)(x)sec(pi/3)=1/cos(pi/3)=1/(1/2)=2$

$∙ x sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $•color(white)(x)sin(pi/3)=sqrt3/2$

$∙ x csc (\frac{π}{3}) = \frac{1}{sin (\frac{π}{3})} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$ $•color(white)(x)csc(pi/3)=1/sin(pi/3)=1/(sqrt3/2)=2/sqrt3$

$∙ x tan (\frac{π}{3}) = \frac{sin (\frac{π}{3})}{cos (\frac{π}{3})} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$ $•color(white)(x)tan(pi/3)=sin(pi/3)/(cos(pi/3))=(sqrt3/2)/(1/2)=sqrt3$

$∙ x cot (\frac{π}{3}) = \frac{1}{tan (\frac{π}{3})} = \frac{1}{\sqrt{3}}$ $•color(white)(x)cot(pi/3)=1/tan(pi/3)=1/sqrt3$

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How do you give the six trigonometric function values of pi/3?

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