SOLUTION: (a) prove(i have done this part already)
cos3a = 4cos^3 a - 3cosa
(b) Show that cos40 is a root of the equation 8x^3 - 6x +1 = 0
Need help with part (b) PLEASE
Algebra ->
Trigonometry-basics
-> SOLUTION: (a) prove(i have done this part already)
cos3a = 4cos^3 a - 3cosa
(b) Show that cos40 is a root of the equation 8x^3 - 6x +1 = 0
Need help with part (b) PLEASE
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You can put this solution on YOUR website! the second part sounds like it's a lot easier than the first part.
all you need to do is find the cosine of 40 degrees and replace x in the equation.
if the value of the equation comes out to be equal to 0, then cosine of 40 degrees is a root of the equation since the roots of the equation are when the value of y is equal to 0.
use your calculator to find cosine of 40 degrees.
that turns out to be equal to be equal to .766044443.
store that in your memory so as not to get rounding errors and use that stored result in place of x in the equation.
when you replace x in your equation with the stored value of .766044443, the result is 0 indicating that cosine of 40 degrees is a root of the equation.
alternately, you can graph your equation and find the roots (when the equation crosses the s-axis are the roots).
the graph of the equation actually crosses the x-axis in 3 places, and the third place (the rightmost intersection) is equal to the cosine of 40 degrees (.76604443).
the graph of the equation is shown below:
i drew a vertical line (well, almost vertical) at x = .766044443 to show you that the graph of the equation crosses the x-axis at that point.
