Need Help finding the period of some trigonomertric functions?

Question

I have an AP Calculus summer assignment.

Here are the questions: (1.) y=4sin(2x)+1
(2.) y=-3tan(pie(x))
(3.) y=cos²(x)
(4.) -3sec(-6x)+2

Answer 1

Hi,

(1.) y=4sin(2x)+1

The normal period for the sine graph is 2π radians or 360°.
When the equation is in the form y = A sin(Bx) + D, divide the normal period by the absolute value of B to get the new period of this function.

In this case, since B = 2, the period is 2π/2 = π radians or 360°/2 = 180°.



(2.) y= -3tan(π(x))

The normal period for the tangent graph is π radians or 180°.
When the equation is in the form y = A tan(Bx) + D, divide the normal period by the absolute value of B to get the new period of this function.

In this case, since B = π, the period is π/π or 1 radian.




(3.) y=cos²(x)

The normal period for the cosine graph is 2π radians or 360°.
When the equation is in the form y = A cos²(Bx) + D, divide the normal period by the absolute value of B to get the new period of this function.

In this case, since B = 1, the period is 2π or 360°.




(4.) -3sec(-6x)+2

The normal period for the secant graph is 2π radians or 360°, just like its reciprocal, the cosine graph.
When the equation is in the form y = A sec(Bx) + D, divide the normal period by the absolute value of B to get the new period of this function.

In this case, since B = -6, the period is 2π/6 = π/3 radians or 360°/6 = 60°


I hope this helps!!

Answer 2

hi

minimum T positive
f(x+T) = f(x)

trig_function(ax) -> T = 2 pi / a

1) T = pi
2) T = 2
3) T = 2 pi
4) T = pi/3

bye

Answer 3

1)2(x+P)=2x+2pi so 2P=2pi and P=pi
2)pi(x+P) =pix+pi so P=1
3)2pi
4) sec=1/cos so 1/cos(-6x)= 1/cos(6x)
6(x+P) =6x+2pi so P = pi/3