Question 492285
Let ( (-2/sqrt(5)), (1/sqrt(5)) ) be the point on the unit circle that corresponds to a real number, t. Find the exact values of the six trigonometric functions of t
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You are working with a right triangle in quadrant II where sin>0 and cos<0.
Opposite side=1/&#8730;5
Adjacent side=-2&#8730;5
Hypotenuse=&#8730;[(1/&#8730;5)^2+(2/&#8730;5)^2]=&#8730;[(1/5)+(4/5)]=&#8730;1=1
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Trig functions:
sin t=(1/&#8730;5)/1=1/&#8730;5
cos t=(-2&#8730;5)/1=-2/&#8730;5=-2&#8730;5/5
tan t=sin t/cos t=(1/&#8730;5)/(-2&#8730;5)= -1/2
csc t=1/sin t=&#8730;5
sec t=1/cos t=-&#8730;5/2
cot t=1/tan t=-2