its so hard. plz help?

Question

12. give the value of:

a. sin ^-1 ( cos x). ( Limit solution to an acute angle)

b. tan ( sin ^-1 x)

Answer 1

a. Use cos x = sqrt(1 - sin^2 x).
b. Use tan x = sin x/cos x.

Answer 2

a. All this problem is asking you to do is give the angle whose sine is equal to the cosine of x. It can be written as arcsin (cos x). Of course the angle is x, but we have to determine what x is. There is only 1 acute angle which fits this bill: (pi/4). Here's why:

sin (pi/4) = (√2)/ 2
cos (pi/4) = (√2)/ 2.

Notice this is not true of -(pi/4) because while the sine is negative, the cosine is positive. Nor will it work for ± (3pi/4), the supplements of ± (pi/4), because they aren't acute angles.

So sin^-1(cos x) or arcsin (cos x) = sin^-1[(√2)/2] = sin^-1[cos (pi/4)], where (pi/4) = x.

b. This problem is asking you to give the tangent of the angle whose sine is equal to x. This could also be written tan [arcsin (x)].

Draw a diagram of a right triangle located in the first quadrant of a Cartesian coordinate system with one of the acute angles located at the origin and the right angle on the x-axis to the right of the origin. The second acute angle's vertex will lie somewhere directly above the vertex of the right angle. Let the vertical leg be of length x and the horizontal leg be of length x'. By definition, the sine of the angle at the origin will be the length of the vertical leg divided by the length of the hypotenuse, h:

sin Θ = x/h.

But since sin Θ = x, then h must = 1.

By the Pythagorean Theorem, the length of the horizontal side is:

x' = √(1² - x²) = √(1 - x²).

Also, by definition, the cosine of the angle at the origin is the side adjacent to it, x', divided by the hypotenuse:

cos Θ = [√(1 - x²) / 1] = √(1 - x²).

The tangent is defined as:

tan Θ = sin Θ / cos Θ = x / x'

Substituting in the values we have calculated for x and x', the tangent is:

tan Θ = x / [√(1 - x²)].

So, the tangent of the angle whose sine is x is:

x / √(1 - x²).

Or, to put it in mathematical jargon:

tan [sin^-1 (x)] = x / √(1 - x²). This is your answer.

Answer 3

a) You're looking for y = sin^-1 (cosx). This means siny = cosx. This means y and x are the two acute angles of the same right triangle, and so y = 90-x.

b) y = tan(sin^-1 x) → tan^-1 y = sin^-1 x
→ for the same angle t, tant=y and sint=x. Then x is the height of a right triangle with hypotenuse=1, base = sqrt(1-x²) and base angle = t. So y = tant = x/sqrt(1-x²).