How do you find the remaining trigonometric functions of #theta# given #tantheta=-1/2# and #sintheta>0#?

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How do you find the remaining trigonometric functions of #theta# given #tantheta=-1/2# and #sintheta>0#?

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Answer 1 · 2017-03-15T17:59:47

#tan t = - 1/2 --> cot t = - 2#
#sin^2 t = 1/(1 + cot^2 t) = 1/(1 + 4) = 1/5#
#sin t = +- 1/sqrt5 = = +- sqrt5/5#
Take the positive value, because sin t > 0
#cos^2 t = 1 - sin^2 t = 1 - 1/5 = 4/5#
#cos t = +- 2/sqrt5#
Because tan t < 0, and sin t > o, then cos t < 0, take the negative value.
#sec t = 1/(cos) = - sqrt5/2#
#csc t = 1/(sin) = sqrt5#

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How do you find the remaining trigonometric functions of #theta# given #tantheta=-1/2# and #sintheta>0#?

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