How do you find the five remaining trigonometric function satisfying #sintheta=3/8#, #costheta<0#?

1 Answer
Jun 4, 2017

#cosΘ=(-√55)/8#
#tanΘ=(-3√55)/55#
#cscΘ=8/3#
#secΘ=(-8√55)/55#
#cotΘ=(-√55)/3#

Explanation:

Since #sin# is positive and #cos# is less than 0, it means that this triangle is in quadrant 2.

Use Pythagorean's theorem (#a^2+b^2=c^2#) and plug in your given values: #a=3# and #c=8# and solve.

#3^2+b^2=8^2##9+b^2=64##b^2=55##b=-√55# (it is negative because it is in quadrant 2 and the #cos# is the x value of the triangle)

Now that you have all side lengths, simply plug them in to the remaining trigonometric functions:

#cosΘ=(-√55)/8#
#tanΘ=3/(-√55)=(-3√55)/55#
#cscΘ=8/3#
#secΘ=8/(-√55)=(-8√55)/55#
#cotΘ=(-√55)/3#