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You should know that the values of sin and cos in the third quadrant are negative, hence the values of tan and cot are positive.
You need to use the basic trigonometric formula such that:
`sin^2x + cos^2x = 1 =gt cos x = - sqrt(1 - sin ^2x)`
Plugging sin x =`-3/5` in the formula above yields:
`cos x = - sqrt(1 - (-3/5)^2) =gt cos x = -sqrt(16/25) =gt cos x = -4/5`
You should use the following formula to find tan x:
`1 + tan^2 x = 1/cos^2 x =gt tan^2 x = 1/cos^2 x - 1`
`tan^2 x = 1/(16/25) - 1 =gt tan^2 x = 25/16 - 1` =>`tan^2 x = 9/16`
Since the value of tan x is positive in the third quadrant =>
tan x = `sqrt(9/16)` => tan x = `3/4`
Hence the requested values for cos x and tan x are: cos x = -`4/5` and tan x = `3/4` .
The answer is very easy. According to the question, x lies in the third quarter. In the third quarter, only tan is positive, sin and cos values are negative. You can use the general formula,
sin^2(x)+cos^2(x) = 1
this gives, (-3/5)^2 + cos^2(x) = 1
therefore, cos^2(x) = 1 - 9/25 = 16/25
taking the squareroot, cos(x) = +/- sqrt(16/25) = +/- (4/5)
since x is in the third quarter, cos(x) is negative and equals to -(4/5).
tan(x) = sin(x)/cos(x) = (-3/5) / (-4/5) = 3/4
Hope it is clear for you.
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