Instruction

1

Use the knowledge of plane geometry to Express

**the sine**through to**the sinus**. According to the definition,**the sine**of the th angle in a right triangle is the ratio length opposite side to hypotenuse, and**sine**om – adjacent sides to the hypotenuse. Even the simple knowledge of the Pythagorean theorem will allow you in some cases to quickly find the desired transformation.2

Express

**the sine**through to**the sinus**, using the elementary trigonometric identity, according to which the sum of the squares of these values gives one. Please note that correctly complete the task, you can only if you know in what quarter there is a specific angle, otherwise you will get two possible outcomes – positive and negative sign.3

Remember the formula of the cast, which also allows to carry out the required operation. According to them, if the number of π/2 added to (or subtracted from) the angle a, is formed to

**the sine**of this angle. The same operations with the number of 3π/2 to give**the sine**, taken with a negative sign. Accordingly, in the case that you are working with to**the sine of**ω, then**the sine**will allow you to obtain the addition or subtraction of 3π/2, and its negative value of π/2.4

Use formulas to find

**the sine**or co**sine of**a double angle to Express**the sine**through to**the sinus**. The sine of a double angle is twice the product of the**sine**and co**sine of**this angle and to**the sine of**twice the angle is the difference between the squares for**the sine**and**sine**.5

Pay attention to the possibility of recourse to formulas of sum and difference of

**sine**s and to**the sine**s of the two angles. If you perform operations with angles a and C, then**the sine of**their sum (difference) is the sum (difference) product**of the sine**of these angles s and to**the sine of**s, and to**the sine**of the sum (difference) is the difference (amount) works for**the sine**s and**sine**s angles, respectively.# Advice 2: How to find the cosine in the triangle

Often the geometric (trigonometric) tasks required to find

**the cosine**of the angle in**the triangle**, because**the cosine**of the angle allows to determine the value of the angle.Instruction

1

To find

and?=b?+c?-2*b*c*cos?, where:

a, b, C be the sidelengths of a triangle (or rather their lengths),

? – the angle opposite the side a (its value).

From these equalities easily is cos?:

cos?=( b?+c?-huh? )/(2*b*c)

Example 1.

There is a triangle with sides a, b, C equal 3, 4, 5 mm respectively.

Find

Solution:

According to the problem conditions we have:

a=3,

b=4,

C=5.

We denote the opposite side and the angle across the? then, according to the formula derived above, we have:

cos?=(b?+c?-huh? )/(2*b*c)=(4?+5?-3?)/(2*4*5)=(16+25-9)/40=32/40=0,8

The answer of 0.8.

**the cosine**of an angle in**a triangle**, the lengths of the sides are known, we can use the theorem**of the cosine of**s. According to this theorem, the squared length of an arbitrary side of a triangle equals the sum of the squares of its two other sides without twice the product of the lengths of these sides into**the cosine**of the angle between them:and?=b?+c?-2*b*c*cos?, where:

a, b, C be the sidelengths of a triangle (or rather their lengths),

? – the angle opposite the side a (its value).

From these equalities easily is cos?:

cos?=( b?+c?-huh? )/(2*b*c)

Example 1.

There is a triangle with sides a, b, C equal 3, 4, 5 mm respectively.

Find

**the cosine**of the angle between the long sides.Solution:

According to the problem conditions we have:

a=3,

b=4,

C=5.

We denote the opposite side and the angle across the? then, according to the formula derived above, we have:

cos?=(b?+c?-huh? )/(2*b*c)=(4?+5?-3?)/(2*4*5)=(16+25-9)/40=32/40=0,8

The answer of 0.8.

2

If the triangle is rectangular, then to find

Suppose you have a rectangular triangle with sides a, b, C, where C is the hypotenuse.

Consider all the options:

Example 2.

Find cos?, if you know the lengths of the sides a and b (sides of triangle)

We use advanced Pythagorean theorem:

c?=b?+huh?,

C=v(b?+huh?)

cos?=(b?+c?-huh? )/(2*b*c)=(b?+b?+huh?-huh?)/(2*b*v(b?+a?))=(2*b?)/(2*b*v(b?+a?))=b/v(b?+huh?)

To check the correctness of the formula, substitute in the values from example 1, i.e.

a=3,

b=4.

Doing elementary calculations, we get:

cos?=0,8.

**the cosine of**a angle is enough to know only the lengths of any two sides of (**the cosine**of a right angle is equal to 0).Suppose you have a rectangular triangle with sides a, b, C, where C is the hypotenuse.

Consider all the options:

Example 2.

Find cos?, if you know the lengths of the sides a and b (sides of triangle)

We use advanced Pythagorean theorem:

c?=b?+huh?,

C=v(b?+huh?)

cos?=(b?+c?-huh? )/(2*b*c)=(b?+b?+huh?-huh?)/(2*b*v(b?+a?))=(2*b?)/(2*b*v(b?+a?))=b/v(b?+huh?)

To check the correctness of the formula, substitute in the values from example 1, i.e.

a=3,

b=4.

Doing elementary calculations, we get:

cos?=0,8.

3

Similarly, is

Example 3.

Known a and C (hypotenuse and opposite side), find cos?

b?=with?-huh?,

b=v(c?-huh?)

cos?=(b?+c?-huh? )/(2*b*c)=(C?-a?+with?-huh?)/(2*s*v (? -a?))=(2*s?-2*a?)/(2*s*v (? -a?))=v (? -huh?)/C.

Substituting the values a=3 and C=5 from the first example, we get:

cos?=0,8.

**the cosine of**the rectangular**triangle**in other cases:Example 3.

Known a and C (hypotenuse and opposite side), find cos?

b?=with?-huh?,

b=v(c?-huh?)

cos?=(b?+c?-huh? )/(2*b*c)=(C?-a?+with?-huh?)/(2*s*v (? -a?))=(2*s?-2*a?)/(2*s*v (? -a?))=v (? -huh?)/C.

Substituting the values a=3 and C=5 from the first example, we get:

cos?=0,8.

4

Example 4.

Famous b and C (the hypotenuse and adjacent side).

Find cos?

Producing the same (shown in examples 2 and 3 of the transition), we obtain that in this case

cos?=b/C.

The simplicity of obtained formula is explained simple: in fact, adjacent to the corner ? side is the projection of the hypotenuse, so its length is equal to the length of the hypotenuse multiplied by cos?.

Substituting the values b=4 and C=5 from the first example, will get:

cos?=0,8

So all our formulas are correct.

Famous b and C (the hypotenuse and adjacent side).

Find cos?

Producing the same (shown in examples 2 and 3 of the transition), we obtain that in this case

**the cosine**in**the triangle**is calculated by a simple formula:cos?=b/C.

The simplicity of obtained formula is explained simple: in fact, adjacent to the corner ? side is the projection of the hypotenuse, so its length is equal to the length of the hypotenuse multiplied by cos?.

Substituting the values b=4 and C=5 from the first example, will get:

cos?=0,8

So all our formulas are correct.

# Advice 3: How to find the cosine, the sine of knowing

In order to obtain a formula linking

**the sine**and co**sine**of the angle, it is necessary to give or to recall some definitions. So,**the sine**of an angle is the ratio (quotient of the) opposite side of a right triangle to the hypotenuse. To**the sine**of an angle is the ratio adjacent side to the hypotenuse.Instruction

1

Draw a right triangle ABC, where angle ABC is a straight line (Fig.1). Consider the ratio of

sin CAB=BC/AC cos CAB=AB/AC.

**the sine**and co**sine**and angle CAB. According to the above definitionsin CAB=BC/AC cos CAB=AB/AC.

2

Remember the Pythagorean theorem - AB^2 + BC^2 = AC^2, where ^2 is the operation of squaring.

Divide left and right side of the equation by the square of the hypotenuse AC. Then the previous equation will look like this:

AB^2/AC^2 + BC^2 AC^2 = 1.

Divide left and right side of the equation by the square of the hypotenuse AC. Then the previous equation will look like this:

AB^2/AC^2 + BC^2 AC^2 = 1.

3

For convenience, we rewrite the equation obtained in step 2, as follows:

(AB/AC)^2 + (BC/AC)^2 = 1.

According to the definitions given in step 1, we get:

cos^2(CAB) + sin^2(CAB) = 1, i.e.

cos(CAB)=SQRT(1-sin^2(CAB)), where SQRT is the operation of taking the square root.

(AB/AC)^2 + (BC/AC)^2 = 1.

According to the definitions given in step 1, we get:

cos^2(CAB) + sin^2(CAB) = 1, i.e.

cos(CAB)=SQRT(1-sin^2(CAB)), where SQRT is the operation of taking the square root.

Useful advice

The magnitude of the sine and cosine of any angle cannot be greater than 1.

# Advice 4: How to find the cosine if you know the sine

**The sine**and

**cosine**is the direct trigonometric functions, for which there are several definitions through a circle in a Cartesian coordinate system, using the solution of a differential equation, using acute angles in a right triangle. Each of these definitions allows us to derive a relationship between these two functions. The following is probably the most simple way to Express

**the cosine**in terms of sine, through their definitions for acute angles of a right triangle.

Instruction

1

Express the sine of an acute angle of a right triangle using the lengths of the sides of this figure. According to the definition, the sine of the angle (α) should be equal to the ratio of the length of the sides (a), lying in front of him - the side to side length (c) opposite the right angle is the hypotenuse: sin(α) = a/c.

2

Find a similar formula for

**the cosine of**a same angle. By definition, this value should be expressed by the ratio of the length of the sides (b) adjacent to this corner (second side) to the length of the side (c), which lies opposite the right angle: cos(a) = a/c.3

Rewrite the equation resulting from the Pythagorean theorem, so that it was involved in the relationship between the legs and the hypotenuse, derived from the previous two steps. To do this, first divide both sides of the original equation of this theorem (a2 + b2 = c2) on the square of the hypotenuse (a2/c2 + b2/c2 = 1), and then the obtained equality rewrite in this form: (a/c)2 + (b/c)2 = 1.

4

Replace in the resulting expression of the ratio of the lengths of the legs and hypotenuse trigonometric functions based on the formulas of the first and second step: sin2(a) + cos2(a) = 1. Express

**the cosine**of the obtained equality: cos(a) = √(1 - sin2(a)). This problem can be considered solved in General.5

If in addition to the common solutions you need to obtain a numerical result, use, for example, a calculator built into the Windows operating system. The link to his startup, locate the subsection "Standard" under "All programs" on the main menu OS. This link is articulated succinctly - "Calculator". To be able to calculate using this program trigonometric functions turn it on "engineering" the interface is press Alt + 2.

6

Enter this in terms of the value of the sine of the angle and click interface designation x2 - so you put up the original value in the square. Then enter on the keypad *-1, hit enter, then type +1 and hit Enter again - this way you subtract one square of sine. Click the button with the icon of the radical, to extract the square root and get the final result.

# Advice 5: What is the sine and cosine

The study of triangles being mathematicians for several millennia. The science of triangles - trigonometry - uses special values: sine and cosine.

## Right triangle

Initially, the sine and cosine arose from the need to calculate values in right triangles. It was observed that if the value of the degree measures of the angles in a right triangle do not change, the aspect ratio, no matter how these parties are neither changed in length, remains always the same.

And it was introduced the concepts of sine and cosine. The sine of an acute angle in a right triangle is the ratio of the opposite leg to the hypotenuse, cosine – adjacent to the hypotenuse.

## Cosines and sines

But the cosines and sines can be applied not only in right triangles. To find the value of the obtuse or acute angle, the sides of any triangle, it is enough to apply theorem law of cosines and law of sines.

The theorem of the cosines is quite simple: the squares of the sides of a triangle is equal to the sum of the squares of the other two sides minus twice the product of these sides into the cosine of the angle between them."

There are two interpretations of the theorem of sines: small and extended. According to small: "In a triangle the angles opposite the sides are proportional". This theorem often extend through the properties of a triangle circumscribed about the circle: In a triangle the angles opposite the sides are proportional and their ratio is equal to the diameter of the circumscribed circle".

## Derivatives

The derivative is a mathematical tool that indicates how quickly changes in the function of relative change of its argument. Derivatives are used in algebra, geometry, Economics and physics, the number of technical disciplines.

When solving problems you must know the table values of the derivatives of trigonometric functions: sine and cosine. The derivative of sine is cosine, and cosine - sine, but with the sign "minus".

## Application in mathematics

Very often the sines and cosines are used when solving right-angled triangles and problems associated with them.

The convenience of sines and cosines is reflected in the technique. Corners and sides were just judged by the theorems of cosines and sinuses by breaking up the complex shapes and objects with "simple" triangles. Engineers and architects, often dealing with the calculations of the aspect ratio and degree measures, spending a lot of time and effort to calculate the cosines and sines of angles is not tabular.

Then "help" came to the table Bradis containing thousands of values of sines, cosines, tangents and cotangent different angles. In Soviet times, some teachers were forced to teach their charges page tables Bradis by heart.