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Find the exact values of the six trigonometric functions of theta given the point (-4, 5) on the terminal side of theta in standard position.
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You have x = -5 and y = -4 ; solve for "r". r^2 = 5^2 + 4^2 = 41 r = sqrt(41) ?
sin(theta) = y/r = -5/sqrt(41) cos(theta) = x/r = -4/sqrt(41) tan(theta) = y/x = 5/4
Make a right triangle with legs 4 and 5. The hypotenuse will be sqrt(41) using pythagoras. Sin = y / hyp = 5 / sqrt(41) cos = x / hyp = 4 / sqrt(41- tan = y / x = 5 / 4 cot = x / y = 4 / 5 sec =hyp / x = sqrt(41) / 4 csc = hyp / y = sqrt(41) / 5
yerp
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