The periodicity identities of trigonometric functions tell us that shifting the graph of a trigonometric function by a certain amount results in the same function.
The sinx and cosx functions as well as their respective reciprocals cscx and secx all have a period of 2π, while tanx and cotx have a period of π. So the periodicity identities of the functions are
sin(x)cos(x)tan(x)=sin(x+2π)=cos(x+2π)=tan(x+π)csc(x)sec(x)cot(x)=csc(x+2π)=sec(x+2π)=cot(x+π).
Analogously,
sin(x)cos(x)=−sin(x+π)=−cos(x+π)csc(x)sec(x)=−csc(x+π)=−sec(x+π).
So basically, if we know the value of the function from 0 to 2π for the first 3 functions, we can find the value of the function at any value. More clearly, we can think of the functions as the values of a unit circle.
The above figure shows that the sine and cosine functions repeat every time we go around the unit circle. In fact we can go around 4π,6π,8π,…,2kπ for a positive integer k and still get the same function. So a more general form of the periodicity identity will be:
sin(x)cos(x)tan(x)=sin(x+2kπ)=cos(x+2kπ)=tan(x+kπ)csc(x)sec(x)cot(x)=csc(x+2kπ)=sec(x+2kπ)=cot(x+kπ).
Find the value of 2sin(11π+x).
We have
2sin(11π+x)=2sin(2⋅5π+π+x)=2sin(π+x)=−2sinx. □
Find the value of 2cos(319π)+sin(27π).
We have
2cos(319π)+sin(27π)=2cos(6π+3π)+sin(3π+2π)=2cos(3π)+sin(2π+π+2π)=2cos(3π)+sin(π+2π)=2cos(3π)−sin(2π)=1−1=0. □
4π
2π
π
2π
Find the fundamental period of f(t)=∣sint∣+∣cost∣.
Note: A function has a period of T if f(t)=f(t+T) for all t. The fundamental period is the smallest positive period.
Find the value of cos2(−8π−x)1−cos2(7π+x).
We have
cos2(−8π−x)1−cos2(7π+x)=cos2(−x)1−cos2(6π+π+x)=cos2x1−cos2(π+x)=cos2x1−(cosx)2=cos2x1−cos2x=cos2xsin2x=tan2x. □