Loading...

PPT – Trigonometry PowerPoint presentation | free to view - id: 11e3e0-ZTc5N

The Adobe Flash plugin is needed to view this content

Trigonometry

A mathematics PowerPoint by Eric Zhao

Trigonometry is the study and solution of

Triangles. Solving a triangle means finding the

value of each of its sides and angles. The

following terminology and tactics will be

important in the solving of triangles.

Pythagorean Theorem (a2b2c2). Only for right

angle triangles Sine (sin), Cosecant (csc or

sin-1) Cosine (cos), Secant (sec or

cos-1) Tangent (tan), Cotangent (cot or

tan-1) Right/Oblique triangle

Part 1. Trigonometric Functions And Solving

Right Triangles

A trigonometric function is a ratio of certain

parts of a triangle. The names of these ratios

are The sine, cosine, tangent, cosecant, secant,

cotangent. Let us look at this triangle

Given the assigned letters to the sides and

angles, we can determine the following

trigonometric functions.

The Cosecant is the inversion of the sine, the

secant is the inversion of the cosine, the

cotangent is the inversion of the tangent.

With this, we can find the sine of the value of

angle A by dividing side a by side c. In order

to find the angle itself, we must take the sine

of the angle and invert it (in other words, find

the cosecant of the sine of the angle).

Try finding the angles of the following triangle

from the side lengths using the trigonometric

ratios from the previous slide.

Click for the Answer

The first step is to use the trigonometric

functions on angle A. Sin ? 6/10 Sin ?

0.6 Csc0.636.9 Angle A36.9 Because all angles

add up to 180, B90-11.53753.1

The measurements have changed. Find side BA and

side AC

Sin342/BA 0.5592/BA 0.559BA2 BA2/0.559 BA3.57

8

The Pythagorean theorem when used in this

triangle states that BC2AC2AB2 AC2AB2-BC2 AC2

12.802-48.802 AC8.8020.53

Part 2 Solving Oblique Triangles

When solving oblique triangles, simply using

trigonometric functions is not enough. You need

The Law of Sines

The Law of Cosines

a2b2c2-2bc cosA b2a2c2-2ac cosB c2a2b2-2ab

cosC

It is useful to memorize these laws. They can be

used to solve any triangle if enough measurements

are given.

REMEMBER

When solving a triangle, you must remember to

choose the correct law to solve it with. Whenever

possible, the law of sines should be used.

Remember that at least one angle measurement must

be given in order to use the law of sines. The

law of cosines in much more difficult and time

consuming method than the law of sines and is

harder to memorize. This law, however, is the

only way to solve a triangle in which all sides

but no angles are given. Only triangles with all

sides, an angle and two sides, or a side and two

angles given can be solved.

Solve this triangle

Click for answers

Because this triangle has an angle given, we can

use the law of sines to solve it. a/sin A b/sin

B c/sin C and subsitute 4/sin28º b/sin B

6/C. Because we know nothing about b/sin B, lets

start with 4/sin28º and use it to solve 6/sin C.

Cross-multiply those ratios 4sin C 6sin

28, divide 4 sin C (6sin28)/4. 6sin282.817.

Divide that by four 0.704. This means that

sin C0.704. Find the Csc of 0.704 º. Csc0.704º

44.749. Angle C is about 44.749º. Angle B is

about 180-44.749-2817.251. The last side is b.

a/sinA b/sinB, 4/sin28º b/sin17.251º,

4sin17.251sin28b, (4sin17.251)/sin28b.

b2.53.

Solve this triangle

Hint use the law of cosines

Start with the law of cosines because there are

no angles given. a2b2c2-2bc cosA. Substitute

values. 2.423.525.22-2(3.5)(5.2)

cosA, 5.76-12.25-27.04-2(3.5)(5.2) cos A,

33.5336.4cosA, 33.53/36.4cos A, 0.921cos A,

A67.07. Now for B. b2a2c2-2ac cosB,

(3.5)2(2.4)2(5.2)2-2(2.4)(5.2) cosB,

12.255.7627.04-24.96 cos B. 12.255.7627.04-24.

96 cos B, 12.25-5.76-27.04-24.96 cos B.

20.54/24.96cos B. 0.823cos B. B34.61. C180-34

.61-67.0778.32.

Part 3 Trigonometric Identities

Trigonometric identities are ratios and

relationships between certain trigonometric

functions. In the following few slides, you will

learn about different trigonometric identities

that take place in each trigonometric function.

Cofunctions

What is the sine of 60º? 0.866. What is the

cosine of 30º? 0.866. If you look at the name of

cosine, you can actually see that it is the

cofunction of the sine (co-sine). The cotangent

is the cofunction of the tangent (co-tangent),

and the cosecant is the cofunction of the secant

(co-secant). Sine60ºCosine30º Secant60ºCosecant3

0º tangent30ºcotangent60º

Other useful trigonometric identities

The following trigonometric identities are useful

to remember.

Sin ?1/csc ? Cos ?1/sec ? Tan ?1/cot ? Csc

?1/sin ? Sec ?1/cos ? Tan ?1/cot ?

(sin ?)2 (cos ?)21 1(tan ?)2(sec ?)2 1(cot

?)2(csc ?)2

Part 4 Degrees and Radians

Degrees and pi radians are two methods of showing

trigonometric info. To convert between them, use

the following equation. 2p radians 360

degrees 1p radians 180 degrees

Convert 500 degrees into radians. 2p radians

360 degrees, 1 degree 1p radians/180, 500

degrees p radians/180 500 500 degrees 25p

radians/9

Part 5 Review

Review 1

Write out the each of the trigonometric functions

(sin, cos, and tan) of the following degrees to

the hundredth place. (In degrees mode). Note

you do not have to do all of them ?

Review 2 a

Solve the following right triangles with the

dimensions given

Review 2 b

Solve the following oblique triangles with the

dimensions given

Review 3

Find each sine, cosecant, secant, and cotangent

using different trigonometric identities to the

hundredth place(dont just use a few identities,

try all of them.).

Review 4 a

Convert to radians

Review 4 b

Convert to degrees

Creator

Director

Producer

Author

Trigonometry

Eric Zhao

Eric Zhao

Eric Zhao

Eric Zhao

MathPower Nine, chapter 6

Basic Mathematics Second edition By Haym Kruglak,

John T. Moore, Ramon Mata-Toledo

Bibliography

The End

(at last)