If #csc theta=4/3#, what is the sin, cos, tan, sec, and cot?

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Answer 1 · 2018-03-27T15:43:28

See below.

Instead of using formulas, it'd be easier to solve it geometrically, with a right triangle.

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Since #csc theta = 1/sintheta = "hypotenuse"/"opposite"=c/a = 4/3#, this means that #a# and #c# are multiples of #3# and #4#, respectively.

In other words, we have #c=4k# and #a=3k#, for a real number #k#.
By the Pythagorean theorem, #b = sqrt(c^2-a^2) = sqrt(16k^2-9k^2) = sqrt(7)*k#.

Finally, for trigonometric functions :

#sin theta = "opposite"/"hypotenuse" = a/c = 3/4#
#cos theta = "adjacent"/"hypotenuse" = b/c = sqrt7/4#

#tan theta = "opposite"/"adjacent" = a/b = 3/sqrt7#
#cot theta = 1/tan theta = b/a = sqrt7/3#

#sec theta = "hypotenuse"/"adjacent" = c/b = 4/sqrt7#.

Answer 2 · 2018-04-02T11:46:05

As below.

#csc theta = 4/3#

#sin theta = 1/csc theta = 1 / (4/3) = 3/4#

#cos^2 theta = 1 - sin^2 theta = 1 - 9 / 16 = 7 / 16#

#cos theta = +- sqrt7 / 4#

#sec theta = 1 / cos theta = +- 4 / sqrt7#

#tan theta = sin theta / cos theta = +- (3/4) / (sqrt7 / 4) = +-3 / sqrt7#

#cot theta = 1 / tan theta = +- sqrt7 / 3#