Cos^4(x) =
(cos²x)² =
(1-sen²x)² =
1² - 2*1*sen²x + sen^4x =
1 - 2sen²x + sen^4x
We are given with an isosceles triangle having a vertex on the curve given y =<span>27-x^2</span> .
The area of the triangle, A= xy = x (27-x^2)
A' = 27-x^2-2x^2 = 0
x = 3
Amax = 3(27-9) = 54 units2
The area of the sector is found by multiplying the area of the circle and the ratio of the angle subtended (measure of the central angle) by the sector to 360.
<h3>How to find the area of a sector?</h3>
1) The formula for area of a sector of a circle is;
A = (θ/360) * πr²
where πr² is area of circle
θ is the angle subtended by the sector
Thus, we conclude that the area of the sector is found by multiplying the area of the circle and the ratio of the angle subtended (measure of the central angle) by the sector to 360.
2) The area of the triangle formed as part of the segment is subtracted from from the area of the sector.
Read more about Area of Sector at; brainly.com/question/22972014
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Amplitude:4
Equation of Midline: 2
Period of function:3
Function shifted left:0.5
Function shifted up: 2
