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Question

Let sin x= 2/5 where "pi"/2 is <=x<="pi" how do i find the other five trigonometric ratios?

Answer 1

You need to know these formulas:
cos = x/r
sin = y/r
tan = y/x
x^2 + y^2 = r^2
In this case, y = 2 and r = 5, so x^2 + 4 = 25
That means x^2 = 21
Since you're in the second quadrant (pi/2 < x < pi ... and, by the way, you'd normally use theta rather than "x" here), the x-value must be negative. So x < -SQR(21)
That means ...
cos = -SQR(21) / 5
tan = 2 / -SQR(21) or -2SQR(21) / 21

The other functions are the reciprocals of these
sec = 1/cos = 5 / -SQR(21) or -5SQR(21) / 21
csc = 1/sin = 5/2
cot = 1/tan = -SQR(21) / 2

Answer 2

You need to do two things.
First, draw a right triangle and label the angle x.
Sin x = opposite side/hypotenuse=2/5
So the sides of the triangle are 2,square root 21, and 5.
From these sides you can find all of the trigonometric ratios.

The next step is to find out if the values are negative or positive
on the interval you have defined.
To do this, draw a graph of a sine and cosine function.
Remember that the sine starts at (0,0), increases to one at pi/2,
comes back down to the x axis at pi, then goes negative.
The cos starts at (0,1), decreases to (pi/2,0), then goes to
(-1,pi) before increasing again.
Also remember that tanx=sin x/cos x.
From this you should be able to determine all of the ratios.
Hope this is helpful.