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.

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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

Sin A

To eliminate the parameter in the given parametric equations x = t^2 + 3 and y = 2t, we can substitute the expression for t from the second equation into the first equation:

x = (y/2)^2 + 3

Now, we can simplify and write this in rectangular form:

x = (y^2/4) + 3

So, the rectangular equation for the given parametric equations is x = (y^2/4) + 3.

Next, let's determine the exact value of each expression without using a calculator.

1. cos(sin^(-1)(1/2)):
To find cos(sin^(-1)(1/2)), we can use the fact that sin^(-1)(x) gives us the angle whose sine is x.
sin^(-1)(1/2) = pi/6 (Since sine of pi/6 is 1/2)
So, cos(sin^(-1)(1/2)) = cos(pi/6) = 1/2 * sqrt(3).

2. sin^(-1)(sin(7pi/6)):
To determine the exact value of sin^(-1)(sin(7pi/6)), we need to take into account the range of the arcsine function, which is defined as the inverse of sine in the interval from -pi/2 to pi/2.
Since 7pi/6 is not within this interval, we can use the identity sin(x) = sin(x + 2pi*n) to find an equivalent angle within the desired range.
sin(7pi/6) = sin(pi/6)
Therefore, sin^(-1)(sin(7pi/6)) = pi/6.