Trigonometric formulas: cosine, sine and tangent of double angle. Trigonometric formulas: cosine, sine and tangent of a double angle Derivation of the formula for the sine of a double angle

In trigonometry, many formulas are easier to derive than to memorize. Cosine of double angle is a wonderful formula! It allows you to obtain formulas for reducing degrees and formulas for half angles.

So, we need the cosine of the double angle and the trigonometric unit:

They are even similar: in the double angle cosine formula it is the difference between the squares of the cosine and sine, and in the trigonometric unit it is their sum. If we express the cosine from the trigonometric unit:

and substitute it into the cosine of the double angle, we get:

This is another double angle cosine formula:

This formula is the key to obtaining the reduction formula:

So, the formula for reducing the degree of sine is:

If in it the alpha angle is replaced by a half angle alpha in half, and the double angle two alpha is replaced by an alpha angle, then we obtain the half angle formula for sine:

Now we can express the sine from the trigonometric unit:

Let's substitute this expression into the double angle cosine formula:

We got another formula for the cosine of a double angle:

This formula is the key to finding the formula for reducing the power of cosine and the half angle for cosine.

Thus, the formula for reducing the degree of cosine is:

If we replace α with α/2, and 2α with α, we obtain the formula for the half argument for the cosine:

Since tangent is the ratio of sine to cosine, the formula for tangent is:

Cotangent is the ratio of cosine to sine. Therefore, the formula for cotangent is:

Of course, in the process of simplifying trigonometric expressions, there is no point in deriving the formula for half an angle or reducing a degree every time. It is much easier to put a sheet of paper with formulas in front of you. And simplification will move faster, and visual memory will turn on memorization.

But it’s still worth deriving these formulas several times. Then you will be absolutely sure that during the exam, when it is not possible to use a cheat sheet, you will easily get them if the need arises.

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Addition formulas allow you to express sin(2*a), cos(2*a) and tan(a) through trigonometric functions of angle a.

1. cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b).

2. sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b).

3. tg(a+b) = (tg(a) +tg(b))/(1-tg(a)*tg(b)).

Let us put a = b in these formulas. As a result, we obtain the following identities:

1. sin(2*a) = 2*sin(a)*cos(a).

2. cos(2*a) = (cos(a)) 2 - (sin(a)) 2 .

3. tg(2*a) = (2*tg(a))/(1-(tg(a)) 2).

These identities are called double angle formulas. Let's look at several examples of using double angle formulas.

Example 1. Find the value of sin(2*a), knowing that cos(a) = -0.8 and a is the 3rd quarter angle. Solution:

First let's calculate sin(a). Since angle a is the third quarter, the sine in the third quarter will be negative:

sin(a) = -v(1-(cos(a)) 2) = -v(1-0.64) = -v0.36 = -0.6.

Using the double angle sine formula we have:

sin(2*a) = 2*sin(a)*cos(a) = 2*sin(a)*cos(a) = 2*(-0.6)*(-0.8) = 0.96 .

Answer: sin(2*a) = 0.96.

Example 2. Simplify the expression sin(a)*(cos(a)) 3 - (sin(a)) 3 *cos(a). Solution:

Let's take sin(a)*cos(a) out of brackets. We get:

sin(a)*(cos(a)) 3 - (sin(a)) 3 *cos(a) = sin(a)*cos(a)*(cos(a)) 2 - (sin(a)) 2).

Now let's use the double angle formulas:

= (1/2)*(2*sin(a)*cos(a))*cos(2*a) = (1/2)*sin(2*a)*sin(2*a) = (1 /4)*sin(4*a).

Answer: sin(a)*(cos(a)) 3 - (sin(a)) 3 *cos(a) = (1/4)*sin(4*a).

Using the double angle formulas you can obtain the following expressions

1 - cos(2*a) = 2*(sin(a)) 2 ,

1 + cos(2*a) = 2*(cos(a)) 2 .

Sometimes when solving examples it is very convenient to use these formulas. Consider the following example:

Example 3. Simplify the expression (1-cos(a))/(1+cos(a)). Solution:

Let's apply the formulas written above for the expressions (1-cos(a)) and (1+cos(a)). To do this, we first represent angle a in the form of the following product 2*(a/2).

As a result of the transformations we get:

(1-cos(a))/(1+cos(a)) = (2*(sin(a/2)) 2)/(2*(cos(a/2)) 2),

Using the definition of tangent we have:

(2*(sin(a/2)) 2)/(2*(cos(a/2)) 2)= (tg(a/2)) 2 .

Answer: (1-cos(a))/(1+cos(a))= (tg(a/2)) 2 .

– there will certainly be tasks on trigonometry. Trigonometry is often disliked for the need to cram a huge number of difficult formulas, teeming with sines, cosines, tangents and cotangents. The site already once gave advice on how to remember a forgotten formula, using the example of the Euler and Peel formulas.

And in this article we will try to show that it is enough to firmly know only five simple trigonometric formulas, and have a general understanding of the rest and derive them as you go. It’s like with DNA: the molecule does not store the complete blueprints of a finished living being. Rather, it contains instructions for assembling it from available amino acids. So in trigonometry, knowing some general principles, we will get all the necessary formulas from a small set of those that must be kept in mind.

We will rely on the following formulas:

From the formulas for sine and cosine sums, knowing about the parity of the cosine function and the oddness of the sine function, substituting -b instead of b, we obtain formulas for differences:

1. Sine of the difference: sin(a-b) = sinacos(-b)+cosasin(-b) = sinacosb-cosasinb
2. Cosine of the difference: cos(a-b) = cosacos(-b)-sinasin(-b) = cosacosb+sinasinb

Putting a = b into the same formulas, we obtain the formulas for sine and cosine of double angles:

1. Sine of double angle: sin2a = sin(a+a) = sinacosa+cosasina = 2sinacosa
2. Cosine of double angle: cos2a = cos(a+a) = cosacosa-sinasina = cos2 a-sin2 a

The formulas for other multiple angles are obtained similarly:

1. Sine of a triple angle: sin3a = sin(2a+a) = sin2acosa+cos2asina = (2sinacosa)cosa+(cos2 a-sin2 a)sina = 2sinacos2 a+sinacos2 a-sin 3 a = 3 sinacos2 a-sin 3 a = 3 sina(1-sin2 a)-sin 3 a = 3 sina-4sin 3a
2. Cosine of triple angle: cos3a = cos(2a+a) = cos2acosa-sin2asina = (cos2 a-sin2 a)cosa-(2sinacosa)sina = cos 3 a- sin2 acosa-2sin2 acosa = cos 3 a-3 sin2 acosa = cos 3 a-3(1- cos2 a)cosa = 4cos 3 a-3 cosa

Before we move on, let's look at one problem.
Given: the angle is acute.
Find its cosine if
Solution given by one student:
Because , That sina= 3,a cosa = 4.
(From math humor)

So, the definition of tangent relates this function to both sine and cosine. But you can get a formula that relates the tangent only to the cosine. To derive it, we take the main trigonometric identity: sin 2 a+cos 2 a= 1 and divide it by cos 2 a. We get:

So the solution to this problem would be:

(Since the angle is acute, when extracting the root, the + sign is taken)

The formula for the tangent of a sum is another one that is difficult to remember. Let's output it like this:

Immediately displayed and

From the cosine formula for a double angle, you can obtain the sine and cosine formulas for half angles. To do this, to the left side of the double angle cosine formula:
cos2 a = cos 2 a-sin 2 a
we add one, and to the right - a trigonometric unit, i.e. the sum of the squares of sine and cosine.
cos2a+1 = cos2 a-sin2 a+cos2 a+sin2 a
2cos 2 a = cos2 a+1
Expressing cosa through cos2 a and performing a change of variables, we get:

The sign is taken depending on the quadrant.

Similarly, subtracting one from the left side of the equality and the sum of the squares of the sine and cosine from the right, we get:
cos2a-1 = cos2 a-sin2 a-cos2 a-sin2 a
2sin 2 a = 1-cos2 a

And finally, to convert the sum of trigonometric functions into a product, we use the following technique. Let's say we need to represent the sum of sines as a product sina+sinb. Let's introduce variables x and y such that a = x+y, b+x-y. Then
sina+sinb = sin(x+y)+ sin(x-y) = sin x cos y+ cos x sin y+ sin x cos y- cos x sin y=2 sin x cos y. Let us now express x and y in terms of a and b.

Since a = x+y, b = x-y, then . That's why

You can withdraw immediately

1. Formula for partitioning products of sine and cosine V amount: sinacosb = 0.5(sin(a+b)+sin(a-b))

We recommend that you practice and derive formulas on your own for converting the difference of sines and the sum and difference of cosines into the product, as well as for dividing the products of sines and cosines into the sum. Having completed these exercises, you will thoroughly master the skill of deriving trigonometric formulas and will not get lost even in the most difficult test, olympiad or testing.