Saturday
December 20, 2014

Homework Help: math

Posted by Anonymous on Friday, August 3, 2007 at 11:59am.

Eliminate the parameter (What does that mean?) and write a rectangular equation for (could it be [t^2 + 3][2t]?)

x= t^2 + 3
y = 2t

Without a calculator (how can I do that?), determine the exact value of each expression.

cos(Sin^-1 1/2)

Sin^-1 (sin 7pi/6)

x= t^2 + 3
y = 2t

Eliminate t using the last equation:

t = y/2

insert this in the first equation.

-----------------------------------

arcsin(1/2) = pi/6

cos[arcsin(1/2)] = cos(pi/6) =
1/2 sqrt[3]

Note that the arcsin function is defined as the inverse of the sin function in the interval from minus pi/2 to pi/2.

This means that arcsin(sin(x)) = x

if x is between -pi/2 and pi/2

If x is not in this interval you can use:

sin(x) = sin(x + 2 pi n)

and

sin(x) = sin(pi-x)

E.g. sin(7pi/6) = sin(pi - 7/6 pi) =
sin(-pi/6) therefore:

sin(sin(7pi/6)) = -pi/6

sin(7pi/6) = sin()







Correction:

E.g. sin(7pi/6) = sin(pi - 7/6 pi) =
sin(-pi/6) therefore:

arcsin[sin(7pi/6)] = -pi/6

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