Finding Exact Value of Trig Expression

AI Thread Summary
To find the exact values of sec-1(√2) and sin-1(1), one must understand inverse functions. For sec-1(√2), the equation sec(a) = √2 leads to determining the corresponding angle where cos(a) = 1/√2. For sin-1(1), the solution is straightforward as sin(a) = 1 corresponds to the angle π/2. Familiarity with the unit circle and memorizing key angles can greatly assist in solving these types of problems. Mastering these concepts will enhance understanding of trigonometric functions and their inverses.
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Homework Statement



Find the exact value of each expression:
a) sec-1(\sqrt{}2)
b) sin-1(1)

Homework Equations



sec\theta=\stackrel{}{}1/cos\theta

The Attempt at a Solution


I've never learned this, but I am really curious in how it is solved.
Is there a formula for this? Thanks!
 
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These are inverse functions so:

a. For what value of x does sec(x) = sqrt(2)

b. For what value of x does sin(x) = 1?
 
whatisphysics said:

Homework Statement



Find the exact value of each expression:
a) sec-1(\sqrt{}2)
b) sin-1(1)

Homework Equations



sec\theta=\stackrel{}{}1/cos\theta

The Attempt at a Solution


I've never learned this, but I am really curious in how it is solved.
Is there a formula for this? Thanks!
Do you understand inverse functions?

IOW, x = f-1(y) <==> y = f(x)

For example, suppose you were asked to find cos-1(1/2).

Let y = cos-1(1/2).
That is equivalent to 1/2 = cos(y). What angle in the interval [0, \pi] has a cosine of 1/2?
 
Mark44 said:
Do you understand inverse functions?

IOW, x = f-1(y) <==> y = f(x)

For example, suppose you were asked to find cos-1(1/2).

Let y = cos-1(1/2).
That is equivalent to 1/2 = cos(y). What angle in the interval [0, \pi] has a cosine of 1/2?

Should I memorize the circle with all the angles?
And this may sound silly...but on (x,y), which is cos and sin? Is it like (cos, sin) on the circle, or the opposite?
 
On the unit circle, x = cos(t) and y = sin(t).
 
A simple way to look at the problem is let a=\sec^{-1}\sqrt{2} then \sec a=\sqrt{2}. From here it is easy to compute the value a by turning sec into cos and using information about known values of cos.
 
Last edited:
Thank you all for the input! I think I will learn to memorize the circle with all the angles...I'm sure that will help.
 
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