#$&*
mth164
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
five degree polynomial
** **
** **
Find an expression for sin(5theta) as a 5th-degree polynomial in the variable sin(theta)
Sin(5theta) = sin^2(theta)+sin^2(theta) +sin(theta) ?
I have tried and tried to figure out how to do this. It was in the book and I know what the answer is but can’t for the life of me figure out how to get it.
Sin(5θ)= 16sin^5θ - 20sin^3θ + 5sinθ
** **
#$&*
mth 164
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
radical in denominator
** **
** **
sec 15pi/8
I got part of the answer, but I don't know how to get rid of a radical within a another radical in denominator.
I got 2/[sqrt(2 + sqrt2)].....the book reduces it to (2-sqrt2)[sqrt(2+sqrt2)].
** **
It is number 25, chapter 7.5 in sullivan text
@& This is also mentioned in my response to a recent assignment, where the angle was 7 pi / 8. I showed details there as well.
First understand that 1 / sqrt(2) is rationalized by multiplying num. and demon. by sqrt(2), obtaining sqrt(2) / (sqrt(2))^2 = sqrt(2) / 2.
Consider next the expression 1 / (1 + sqrt(2).
If we multiply numerator and denominator by 1 - sqrt(2) the denominator becomes 1 - (sqrt(2))^2 = 1 - 2 = -1. So the square root is now out of the denominator.
The numerator becomes just 1 - sqrt(2) so the result is
(1 - sqrt(2) ) / (-1) = sqrt(2) - 1.
In the problem you ask about the denominator is sqrt(2 + sqrt(2)).
If you multiply numerator and denominoator of your expression by sqrt(2 + sqrt(2)) you end up with
2 sqrt(2 + sqrt(2)) / (2 + sqrt(2) ).
This would be further rationalized by multiplying both num. and denom. by 2 - sqrt(2).*@