trig help.....?

Question

how do you do this??? please show me how you got your answer

write an algebraic expression for the given expression.
- sin(arccos x)

- sec(arctan 3x)

Answer 1

Find sin(arccos x).

Let arccos(x) = u ...(1)
Then cos(u) = x ...(2)

cos^2(u) + sin^2(u) = 1
Therefore:
sin^2(u) = 1 - cos^2(u)

Substituting for cos(u) from (2):
sin^2(u) = 1 - x^2

sin(u) = +/- sqrt(1 - x^2)

Using (1):
sin(arccos(x)) = +/- sqrt(1 - x^2).

To choose the correct sign for the square root, you need to know in which quadrant the angle x lies.


Find sec(arctan 3x).

Let tan(u) = 3x
Then:
sec(arctan(3x)) = sec(u)

1 + tan^2(u) = sec^2(u)
1 + (3x)^2 = sec^2(u)
sec(u) = +/- sqrt[ 1 + (3x)^2 ]
sec(arctan(3x)) = +/- sqrt(1 + 9x^2).

Once again, you need to know the quadrant in which 3x falls to select the right sign.