A Review for Ms. Ma’s Physics 11

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 (a 2 + b 2 = c 2 ). True ONLY for right angle triangles Sine (sin), and its inverse, Cosecant (csc or sin -1 ) Cosine (cos), and its inverse, Secant (sec or cos -1 ) Tangent (tan), and its inverse, Cotangent (cot or tan -1 ) Right or oblique triangle

A trig function is a ratio of certain parts of a triangle. Again, the names of these ratios are: the sine, cosine, tangent, cosecant, secant, and cotangent. Let us look at this triangle … a c b ө A B C Given the assigned letters to the sides and angles, we can determine the following trigonometric functions. Remember this: SOH CAH TOA The cosecant is the inverse of the sine, the secant is the inverse of the cosine, & the cotangent is the inverse 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). Sin θ= Cos θ= Tan θ= Side Opposite Side Adjacent Side Opposite Hypothenuse = = = a b c a b c

Try finding the angles of the following triangle from the side lengths using the trigonometric ratios from the previous slide θ A B C α β A first step is to use the trig functions on ∠ A. Sin θ = 6/10 Sin θ = 0.6 Csc0.6 ~ 36.9 Angle A ~ 36.9° Because all ∠ ’s add up to 180° in a triangle, ∠ B = 90° ° = 53.1° Note: SIG FIGS are WRONG – we’ll review these soon

C 2 34º A B α β The measurements have changed. Find side BA and side AC. Sin34 = 2/BA = 2/BA 0.559BA = 2 BA = 2/0.559 BA ~ (wrong SIG FIGS again) The Pythagorean theorem when used on this triangle states that … BC 2 + AC 2 = AB 2 AC 2 = AB 2 - BC 2 AC 2 = = AC = ~ 3

When solving oblique triangles, using trig functions is not enough. You need … The Law of Sines The Law of Cosines a 2 =b 2 +c 2 -2bc cosA b 2 =a 2 +c 2 -2ac cosB c 2 =a 2 +b 2 -2ab cosC It is useful to memorize these laws. They can be used to solve any triangle if enough measurements are given. a c b A B C

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. You need 3 pieces of information.

a=4 c=6 b A B C 28º Solve this triangle.

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 substitute: 4/sin28º = b/sin B = 6/sin C. Because we know nothing about b/sin B, let’s start with 4/sin28º and use it to solve 6/sin C. Cross-multiply those ratios: 4*sin C = 6*sin 28, then divide by 4: sin C = (6*sin28)/4. a=4 c=6 b A B C 28º

6*sin28= Divide that by four: This means that sin C = Find the Csc (sin -1 ) of º. Csc0.704º = Angle C is about º. Angle B is about 180º º – 28º = º. The last unknown is side is b. a/sinA = b/sinB 4/sin28º = b/sin17.251º 4*sin = sin28*b (4*sin17.251) / sin28 = b  b ~ 2.53

a=2.4 c=5.2 b=3.5 A B C Solve this triangle. Hint: use the law of cosines (as you only have three side lengths, and no angles) a 2 = b 2 + c 2 -2bc cosA. Substitute values = (3.5)(5.2) cosA = -2(3.5)(5.2) cos A = 36.4 cosA / 36.4 = cos A  = cos A, take the secant  A = 67.1

Now for B. b 2 = a 2 + c 2 -2ac cosB (3.5) 2 = (2.4) 2 + (5.2) 2 -2(2.4)(5.2) cosB = cos B = cos B / = cos B = cos B. Take the secant... B = 34.61º C = 180º º º = 78.32º

Whew!

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.

What is the sine of 60º? What is the cosine of 30º? 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). Sine60º=Cosine30º Secant60º=Cosecant30º tangent30º=cotangent60º

Sin θ=1/csc θ Cos θ=1/sec θ Tan θ=1/cot θ Csc θ=1/sin θ Sec θ=1/cos θ Tan θ=1/cot θ The following trigonometric identities are useful to remember. (sin θ) 2 + (cos θ) 2 =1 1+(tan θ) 2 =(sec θ) 2 1+(cot θ) 2 =(csc θ) 2

Degrees and pi radians are two methods of showing trigonometric info. To convert between them, use the following equation. 2π radians = 360 degrees 1π radians= 180 degrees Convert 500 degrees into radians. 2π radians = 360 degrees, 1 degree = 1π radians/180, 500 degrees = π radians/180 * degrees = 25π radians/9

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 1. 45º 2. 38º 3. 22º 4. 18º 5. 95º 6. 63º 7. 90º º º º º º º º º º º º º º º º º º

Solve the following right triangles with the dimensions given 5 c 22 A B C A B C A a c 13 B C 52 ºc 12 8 º A B C

Solve the following oblique triangles with the dimensions given A B C a 25 b 28 º A B C 31 º 15 c º A B C 5 c 8 A B C 168 º

1. 45º 2. 38º 3. 22º 4. 18º 5. 95º 6. 63º 7. 90º º º º º º º º º º º º º º º º º º Find each sine, cosecant, secant, and cotangent using different trigonometric identities to the hundredth place (don’t just use a few identities, try all of them.).

Convert to radians 52º 34º 35º 46º 74º 36º 15º 37º 94º 53º 174º 156º 376º 324º 163º 532º 272º 631º 856º 428º 732º 994º 897º 1768º 2000º

Convert to degrees 3.2π rad 3.1π rad 1.3π rad 7.4π rad 6.7π rad 7.9 rad 5.4π rad 9.6π rad 3.14π rad 6.48π rad 8.23π rad 5.25π rad 72.45π rad 93.16π rad 25.73π rad 79.23π rad π rad π rad π rad π rad

Creator Eric Zhao Director Eric Zhao Producer Eric Zhao Author Eric Zhao MathPower Nine, chapter 6 Basic Mathematics Second edition By Haym Kruglak, John T. Moore, Ramon Mata-Toledo