Trigonometry. Angles add to 180° презентация

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

Right triangles
We only care about right triangles
A right triangle is one in which one of the angles is 90°
Here’s a right triangle:
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
Here’s the
right angle
hypotenuse
adjacent
opposite

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.

The Pythagorean Theorem
If you square the length of the two 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)

5-12-13
There are few triangles with integer 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

Ratios
Since a triangle has three sides, 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
The ratios depend on the shape of the triangle (the angles) but not on the size

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

Using the ratios
With these functions, if you know an angle (in addition to the right angle) and the length of a side, you can compute all other angles and lengths of sides
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

Solve equations:
а) cos(x/5)=1
The 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.

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) 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

The hard part
If you understood this lecture, you’re in great shape for doing all kinds of things with basic graphics
Here’s the part I’ve always found the hardest:
Memorizing the names of the ratios
sin = opp / hyp
cos = adj / hyp
tan = opp / adj