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Trigonometry

Trigonometry. 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.

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Trigonometry

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  1. Trigonometry A Review for Ms. Ma’s Physics 11

  2. 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 (a2 + b2 = c2). 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

  3. Part 1. Trigonometric Functions And Solving Right Triangles

  4. B c a Side Opposite a Sin θ= Cos θ= Tan θ= = Hypothenuse c Side Adjacent b ө = A c Hypothenuse C b Side Opposite a = Side Adjacent b 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 … Given the assigned letters to the sides and angles, we can determine the following trigonometric functions. Remember this: SOHCAHTOA 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).

  5. B 10 α 6 β θ A C 8 Try finding the angles of the following triangle from the side lengths using the trigonometric ratios from the previous slide. A first step is to use the trig functions on ∠A. Sin θ = 6/10 Sinθ = 0.6 Csc0.6 ~ 36.9 Angle A ~ 36.9° Because all ∠’s add up to 180° in a triangle, ∠B = 90° - 36.9° = 53.1° Note: SIG FIGS are WRONG – we’ll review these soon

  6. B α 2 β 34º A C The measurements have changed. Find side BA and side AC. Sin34 = 2/BA 0.559 = 2/BA 0.559BA = 2 BA = 2/0.559 BA ~ 3.578 (wrong SIG FIGS again) The Pythagorean theorem when used on this triangle states that … BC2 + AC2 = AB2 AC2 = AB2 - BC2 AC2 = 12.802 - 4 = 8.802 AC = 8.8020.5 ~ 3

  7. Part 2. Solving Oblique Triangles

  8. B c a A C b When solving oblique triangles, using trig functions is not enough. You need … The Law of Sines The Law of Cosines a2 =b2+c2-2bc cosA b2=a2+c2-2ac cosB c2=a2+b2-2ab cosC It is useful to memorize these laws. They can be used to solve any triangle if enough measurements are given.

  9. 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. You need 3 pieces of information.

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

  11. B c=6 a=4 28º A C b 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.

  12. 6*sin28=2.817. Divide that by four: 0.704. This means that sin C = 0.704. Find the Csc (sin-1) of 0.704 º. Csc0.704º = 44.749. Angle C is about 44.749º. Angle B is about 180º - 44.749º – 28º = 17.251º. The last unknown is side is b. a/sinA = b/sinB 4/sin28º = b/sin17.251º 4*sin17.251 = sin28*b (4*sin17.251) / sin28 = b  b ~ 2.53

  13. B c=5.2 a=2.4 A b=3.5 C Solve this triangle. Hint: use the law of cosines (as you only have three side lengths, and no angles) a2 = b2 + c2 -2bc cosA. Substitute values. 2.42 = 3.52 + 5.22 -2(3.5)(5.2) cosA 5.76 - 12.25 - 27.04 = -2(3.5)(5.2) cos A 33.53 = 36.4 cosA 33.53 / 36.4 = cos A  0.921 = cos A, take the secant  A = 67.1

  14. Now for B. b2 = a2 + c2 -2ac cosB (3.5)2 = (2.4)2 + (5.2)2 -2(2.4)(5.2) cosB 12.25 = 5.76 + 27.04 -24.96 cos B 12.25 - 5.76 - 27.04 = -24.96 cos B 20.54 / 24.96 = cos B 0.823 = cos B. Take the secant ... B = 34.61º C = 180º - 34.61º - 67.07º = 78.32º

  15. Whew!

  16. Part 3 Trigonometric Identities

  17. 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.

  18. 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). Sine60º=Cosine30º Secant60º=Cosecant30º tangent30º=cotangent60º

  19. Other useful trigonometric identities The following trigonometric identities are useful to remember. 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 (sin θ)2 + (cos θ)2=1 1+(tan θ)2=(sec θ)2 1+(cot θ)2=(csc θ)2

  20. Part 4 Degrees and Radians

  21. 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 * 500 500 degrees = 25π radians/9

  22. Part 5 Review

  23. 1. 45º 7. 90º 13. 47º 19. 75º 2. 38º 8. 152º 14. 442º 20. 34º 3. 22º 9. 112º 15. 123º 21. 53º 4. 18º 10. 58º 16. 53º 22. 92º 5. 95º 11. 345º 17. 41º 23. 153º 6. 63º 12. 221º 18. 22º 24. 1000º 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 

  24. B B B B 52 º c a 20 c c 5 9 A C 13 8 º A A A C C C 12 22 18 Review 2 a Solve the following right triangles with the dimensions given

  25. B B B B c c 22 25 31 º 12 5 15 a 168 º 35 º 28 º A A A A 24 14 8 b C C C C Review 2 b Solve the following oblique triangles with the dimensions given

  26. 1. 45º 7. 90º 13. 47º 19. 75º 2. 38º 8. 152º 14. 442º 20. 34º 3. 22º 9. 112º 15. 123º 21. 53º 4. 18º 10. 58º 16. 53º 22. 92º 5. 95º 11. 345º 17. 41º 23. 153º 6. 63º 12. 221º 18. 22º 24. 1000º Review 3 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.).

  27. 34º 15º 156º 272º 994º 52º 36º 174º 532º 732º 35º 37º 376º 631º 897º 46º 94º 324º 856º 1768º 74º 53º 163º 428º 2000º Review 4 a Convert to radians

  28. 52.652π rad 6.7π rad 72.45π rad 3.2π rad 3.14π rad 7.9 rad 93.16π rad 435.96π rad 6.48π rad 3.1π rad 5.4π rad 25.73π rad 14.995π rad 8.23π rad 1.3π rad 9.6π rad 79.23π rad 5.25π rad 745.153π rad 7.4π rad Review 4 b Convert to degrees

  29. Creator Director Producer Author Trigonometry MathPower Nine, chapter 6 Basic Mathematics Second edition By Haym Kruglak, John T. Moore, Ramon Mata-Toledo Eric Zhao Eric Zhao Eric Zhao Eric Zhao Bibliography The End (at last)

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