If you are given a simple expression, in which there is only one trigonometric function (sin, cos, tg, ctg, sec, cosec), and the angle inside the function is not multiplied by any number, and she is not raised to some degree – use the definition. For expressions containing sin, cos, sec, cosec feel free to put a period 2P, and if the equation is tg, ctg – P. for Example, for the function y=2 p+5 period is equal to 2P.
If the angle x is under the sign of the trigonometric function multiplied by any number, then to find the period of this function, divide the standard period for this number. For example, you are given the function y= sin 5x. The standard period for sine – 2P, dividing it into 5, you get 2P/5 – this is the desired period in the expression.
To find the period of trigonometric functions raised to a power, evaluate the parity of the degree. For even degree reduce the standard period of two times. For example, if you are given the function y=3 cos^2x, then the standard period 2P reduced in 2 times, thus the period is equal to P. note that the functions tg, ctg in any degree periodicity P.
If you are given an equation containing the product or quotient of two trigonometric functions, first find the period for each separately. Then find the minimum number that was big enough to contain in itself the whole of both periods. For example, given the function y=tgx*cos5x. For tangent, the period P, the cosine of 5x – period 2P/5. The minimum number that can fit both the period is 2P, therefore, the required period is – 2P.
If you find it difficult to act in the suggested manner or are unsure of the answer, try to act according to the definition. Take as period of the function T, it is greater than zero. Substitute in the equation instead of x the expression (x+T) and solve the resulting equation as if T was a parameter or number. As a result, you will find the value of trigonometric functions and will be able to choose the minimum period. For example, as a result of simplification you have got the identity sin (T/2)=0. The minimum value of T at which it is performed, is equal to 2P, it will be the answer to the problem.