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
1 answer
1 answer