End-of-Course Assessment
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π/2 ÷ 4=π/8 Let's make a table of values to show these five points. We will start with 0.
| 2cos(4x) | f(x)=2 cos(4x) | Point | |
|---|---|---|---|
| x=0 | 2 cos(4* 0) | 2 cos(0)=2 | ( 0,2) |
| x=π/8 | 2 cos (4* π/8) | 2 cos (π/2)= | ( π/8, ) |
| x=π/4 | 2 cos (4* π/4 ) | 2cos(π)=-2 | ( π/4,-2 ) |
| x=3π/8 | 2 cos (4* 3π/8) | 2 cos (3π/2)= | ( 3π/8, ) |
| x=π/2 | 2 cos (4* π/2) | 2cos(2π)=2 | ( π/2,2) |
Notice that the maximums will occur at 2 and the minimums will occur at -2 because the amplitude of the graph is 2. Let's now see these points on the coordinate plane!
Finally, we can connect the points to graph one cycle of the function.
Great! Now, we will also continue the pattern for one more cycle to have it over two periods.
.LHS /4.=.RHS /4.
Rearrange equation
Use a calculator
Round to 2 decimal place(s)