[sin(x)^3 + cos(x)^3] / [sin(x) + cos(x)]
= [(sin(x) + cos(x))(sin(x)^2 - sin(x)cos(x) + cos(x)^2)] / [sin(x) + cos(x)]
***Now you can cancel a "sin(x) + cos(x)" from the top and bottom of the fraction. This makes the bottom of the fraction equal to 1. I am just going to write the next step without a 1 on the bottom of the fraction (x/1=x).
So now you just have:
= (sin(x)^2 - sin(x)cos(x) + cos(x)^2) *I'm going to move some terms around now. ~
Not doing any computation in this step.
= (sin(x)^2 + cos(x)^2 - sin(x)cos(x)) *Now we know that cos(x)^2 + sin(x)^2 = 1.
= 1 - sin(x)cos(x)
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Cosine squared theta = 1 + Sine squared theta
A "zero of a function" is a point where the dependent value (usually, Y) is zero. In the function f(x) = x2 - 2, for example, there are zeroes at -1.414 and +1.414.The zeroes of the sine function are at all integer multiples of pi, i.e. 0, pi, 2pi, 3pi, etc. The zeroes of the cosine function are at the same points plus pi/2, i.e. pi/2, 3pi/2, 5pi/2, etc.Another way to look at this is that the zeroes of sine are the even multiples of pi/2, and the zeros of cosine are the odd multiples of pi/2.
9y cubed plus 2y squared
2 cubed = 2*2*2 =8. 8+8+8= 24
3 cube =27 4 cube =64 5 cube =125 6cube=216 so 3 cubed + 4 cubed + 5 cubed = 6 cubed