Instruction

1

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.

2

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.

3

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.

4

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.

5

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.

# Advice 2 : How to investigate the function of parity

Examination of functions of even parity and odd parity helps to build the graph of the function and to investigate the nature of its behavior. For this study it is necessary to compare the function written for the argument "x" and argument "-x".

Instruction

1

Record function, the study on which it is to be in the form y=y(x).

2

Replace the function argument with "-x". Substitute the argument in the function expression.

3

Simplify the expression.

4

So you got the same function written for the arguments "x" and "-x". Look at these two entries.

If y(-x)=y(x), it is an even function.

If y(-x)=-y(x), then it is an odd function.

If about the function, we cannot say that y(-x)=y(x) or y(-x)=-y(x), the parity property is a function of the General form. That is, it is neither even nor odd.

If y(-x)=y(x), it is an even function.

If y(-x)=-y(x), then it is an odd function.

If about the function, we cannot say that y(-x)=y(x) or y(-x)=-y(x), the parity property is a function of the General form. That is, it is neither even nor odd.

5

Write down your insights. Now you can use them in plotting functions or further analytical study of the properties of the function.

6

To talk about parity and odd functions even in the case where the already set schedule function. For example, a graph was the result of physical experiment.

If the graph of a function symmetric about the y-axis, then y(x) is an even function.

If the graph of a function symmetric about the x-axis, x(y) is an even function. x(y) is the inverse of the function y(x).

If the graph of a function symmetric about the origin (0,0), then y(x) is an odd function. Odd is also the inverse function x(y).

If the graph of a function symmetric about the y-axis, then y(x) is an even function.

If the graph of a function symmetric about the x-axis, x(y) is an even function. x(y) is the inverse of the function y(x).

If the graph of a function symmetric about the origin (0,0), then y(x) is an odd function. Odd is also the inverse function x(y).

7

It is important to remember that the concept of parity and of odd functions has a direct relationship with the domain of the function. If, for example, is even or odd the function exists at x=5, then it does not exist and when x=-5, what can be said about the function of the General form. When establishing parity and odd, pay attention to the domain of the function.

8

Examination of functions of even parity and odd parity correlated with the presence of many function values. To find the set of values of even functions, it is sufficient to consider half of the functions to the right or to the left of zero. If x>0 an even function y(x) takes values from A to b, then the same values it will take and when x<0.

To find the set of values taken by an odd function, is also sufficient to consider only one part of the function. If x>0 odd function y(x) takes a range of values from A to b, when x<0 it will make a symmetrical range of values from (- ) to (-A).

To find the set of values taken by an odd function, is also sufficient to consider only one part of the function. If x>0 odd function y(x) takes a range of values from A to b, when x<0 it will make a symmetrical range of values from (- ) to (-A).

# Advice 3 : How to find the minimum value of the function

The need to find the minimum

**value**of a mathematical**function**is a practical interest in the solution of applied tasks, for example, in the economy. Great**value**for business is to minimize losses.Instruction

1

To find the minimum

**value****of the function**, you need to determine what value of the argument x0 to run the inequality y(x0) ≤ y(x), where x ≠ x0. Usually this problem is solved at a certain interval or in the whole region of values**of the function**, if one is not specified. One aspect of the solution is the location of the stationary points.2

A stationary point is called

**a value**argument, in which the derivative**of the function**vanishes. According to the Fermat's theorem if a differentiable function takes the extreme**value**at some point (in this case a local minimum), then this point is stationary.3

The minimum

**value of**a function often accepts it at this point, however, it is possible to determine not always. Moreover, it is not always possible to say with certainty what is the minimum**of the function**or it takes an infinitely small**value**. Then, as a rule, find the limit to which it tends when descending.4

In order to determine the minimum

**value****of the function**, you need to perform a sequence of actions, consisting of four stages: finding domain**of a function**, obtaining stationary points, analysis values**of the function**at these points and at the ends of the interval, identifying a minimum.5

So, imagine you are given some function y(x) on the interval with boundary points A and B. Find the area of its definition and find out whether the interval is its subset.

6

Calculate the derivative

**of the function**. Paranaita the resulting expression to zero and find the roots of the equation. Check whether these stationary points in the interval. If not, then at the next stage, they are ignored.7

Consider the interval subject to boundary types: open, closed, combined or infinite. It depends on how you look for the minimum

**value**. For example, the interval [A, b] is a closed interval. Substitute them into the function and calculate values. Do the same with a stationary point. Select the minimum result.8

With open and infinite intervals the situation is more complicated. There will have to find one-sided limits that do not always give an unambiguous result. For example, for the interval with one closed and one-punctured boundary of [A, b) should find the function at x = A and one-sided limit lim y as x → -0.