Similar presentations:

# Trigonometry. Angles add to 180°

## 1. Trigonometry

Executed: Bissarinova A.20-Aug-17

## 2. Angles add to 180°

The angles of a triangle always add up to 180°20°

44°

68°

44°

68°

+ 68°

180°

68°

30°

120°

20°

30°

+ 130°

180°

## 3. Right triangles

We only care about right trianglesA right triangle is one in which one of the angles is 90°

Here’s a right triangle:

Here’s the angle

we are looking at

Here’s the

right angle

opposite

adjacent

We call the longest side the hypotenuse

We pick one of the other angles--not the right angle

We name the other two sides relative to that angle

## 4. Example.

Solve the equations:a) cos (x / 5) = 1

Decision: A) This time

we proceed directly to the

calculation of the roots of

the equation at once: x / 5

= ± arccos (1) + 2πk.

Then x / 5 = πk => x = 5πk

The answer is: x = 5πk,

where k is an integer.

b) tg (3x-π / 3) = √3

B) We write in the form:

3x-π / 3 = arctg (√3) +

πk. We know that: arctg

(√3) = π / 3 3x-π / 3 = π /

3 + πk => 3x = 2π / 3 +

πk => x = 2π / 9 + πk / 3

The answer is: x = 2π / 9

+ πk / 3, where k is an

integer.

## 5. The Pythagorean Theorem

If you square the length of thetwo shorter sides and add

them, you get the square of the

length of the hypotenuse

adj2 + opp2 = hyp2

32 + 42 = 52, or 9 + 16 = 25

hyp = sqrt(adj2 + opp2)

5 = sqrt(9 + 16)

## 6. 5-12-13

There are few triangles withinteger sides that satisfy the

Pythagorean formula

3-4-5 and its

multiples (6-8-10, etc.)

are the best known

5-12-13 and its multiples

form another set

25 + 144 = 169

opp

adj

## 7. Ratios

Since a triangle has threesides, there are six ways to

divide the lengths of the sides

Each of these six ratios has a

name (and an abbreviation)

Three ratios are most used:

sine = sin = opp / hyp

cosine = cos = adj / hyp

tangent = tan = opp / adj

The other three ratios are

redundant with these and can

be ignored

opposite

opposite

Ratios

adjacent

The ratios depend on the

shape of the triangle (the

angles) but not on the size

adjacent

## 8. Example.

Solve the equations:cos (4x) = √2 / 2.

And find all the roots on the

interval [0; Π].

Decision: Let us solve our equation

in general form: 4x = ± arccos (√2 /

2) + 2πk 4x = ± π / 4 + 2πk;

X = ± π / 16 + πk / 2;

Now let's see what roots get into

our segment.

When k <0, the solution is also less

than zero, we do not fall into our

segment.

For k = 0, x = π / 16, we are in the

given interval [0; Π].

For k = 1, x = π / 16 + π / 2 = 9π / 16,

again we have got.

For k = 2, x = π / 16 + π = 17π / 16,

and here we are no longer there, and

therefore for large k we also will not

fall.

The answer is: x = π / 16, x = 9π / 16

## 9. Using the ratios

With these functions, if you know an angle (in addition to theright angle) and the length of a side, you can compute all other

angles and lengths of sides

opposite

adjacent

If you know the angle marked in red (call it A) and you know

the length of the adjacent side, then

tan A = opp / adj, so length of opposite side is given by

opp = adj * tan A

cos A = adj / hyp, so length of hypotenuse is given by

hyp = adj / cos A

## 10. Solve equations:

а) cos(x/5)=1The answer is: x = 5πk,

where k is an integer.

б)tg(3x- π/3)= √3

The answer is: x = 2π / 9

+ πk / 3, where k is an

integer.

## 11.

Decision: A) This time weproceed directly to the

calculation of the roots of the

equation at once:

x / 5 = ± arccos (1) + 2πk.

Then x / 5 = πk => x = 5πk

The answer is: x = 5πk,

where k is an integer.

B) We write in the form: 3x-π

/ 3 = arctg (√3) + πk.

We know that:

arctg (√3) = π / 3 3x-π / 3 = π

/ 3 + πk => 3x = 2π / 3 + πk

=> x = 2π / 9 + πk / 3

The answer is:

x = 2π / 9 + πk / 3, where k is

an integer

## 12. The hard part

If you understood this lecture, you’re in great shape fordoing all kinds of things with basic graphics

Here’s the part I’ve always found the hardest:

sin = opp / hyp

cos = adj / hyp

tan = opp / adj

adjacent

opposite

Memorizing the names of the ratios