Compute exactly: sec[cos^-1 (sqrt 2/2) + sin^-1 (-1)]?

Question

I am having a hard time with my trigonometry homework, if anyone can help me with the above problem, I will be very appreciative! The notations ^-1 after the cos and the sin stand for the inverse. I wasn't sure how to notate that. Thank you again for any help you can give!!!!

Answer 1

I'm only going to calculate the principal values.

Let x = cos^[-1](sqrt(2) / 2).
Therefore, cos(x) = sqrt(2) / 2.
So, x = pi / 4 radians.

Let y = sin[-1](-1).
Therefore, sin(y) = -1.
So, y = - pi / 2 radians.

Your expression is then : sec(x + y)
= sec(pi / 4 - pi / 2)
= sec(- pi / 4)
= 1 / cos(- pi / 4)
= 1 / [sqrt(2) / 2]
= 2 / sqrt(2)
= sqrt(2)

Answer 2

cos ^ (sqrt(2)/2 = +/-pi/4 because cos pi/4 and cos -pi/4 = sqrt(2)/2

sin ^-(-1) = 3pi/2

now add the 2 we get (3pi/2+pi/4) or (3[pi/2) - pi/4
let x be the sum
taking sec of both sides we get sec (x) = sqrt(2)

Answer 3

sec[cos^-1 (sqrt 2/2) + sin^-1 (-1)]
sec(π/4+3π/2)=sec(7π/4)=√2

Answer 4

Maybe show some work, and we can help you where you're having trouble.