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§9.1 Trigonometric Identities sin (θ) = y and sin (-θ) = r so, sin (-θ) = - sin (θ) and cos (θ) = x and cos (-θ) = r so, cos (-θ) = cos (θ) r y x θ x -y -θ r -y r x r And, Tan (-θ) = sin (-θ) = cos (-θ) - sin (θ) = – sin (θ) = - tan (θ) cos (θ) cos (θ) Note: You may have learned that sine is an odd f(n) [it’s symmetric about the orginin so if (x, y), then (-x, -y)] and cosine is an even f(n) [it’s symmetric about the y-axis so if (x, y) then (-x, y)] Fundamental Identities Reciprocal Identities cot θ = 1 tan θ csc = 1 sec θ = sin θ 1 cos θ Quotient Identities tan θ = sin θ cos θ cot θ = cos θ sin θ Pythagorean Identities sin2 θ + cos2 θ = 1 1 + cot2 θ = csc2 θ tan2 θ + 1 = sec2 θ sin (-θ) = -sin (θ) cos (-θ) = cos θ tan (-θ) = -tan θ csc (-θ) = -csc θ sec (-θ) = sec θ cot (-θ) = -cot θ Negative Angle Identities Sometimes it will be necessary to re-express one function in terms of another. This can be useful in graphing and in differentiation and integration in Calculus. To do this, try the following: 1) Find an identity that relates the functions 2) Solve for the function of desire Example: Express tan θ in terms of cos θ Now, we’ll do a little more of the same, but try to use only sin and cos. This is an important skill to possess, hence while we’ll practice this skill by itself in this section. Here’s a strategy: 1) Use sin & cos to re-write the function(s) 2) Simplify using algebra skills Example: Write (sec θ + csc θ) in terms of sine and cosine and simplify Your Turn Example: Write cos x in terms of csc x. (Hint: Think Pythagorean Id. & reciprocate) Here are some of the hints that Rockswold, Lial & Hornsby have for verifying Trig Identities in their book, A Graphical Approach to Precalculus with Limits: A Unit Circle Approach, 5th Ed., p. 633. We will work on verifying identities first using only one side of the equation. The purpose in doing this is to prepare for the use of identities in solving Calculus problems. We need skills of rewriting just one side that we will practice this time and also the skills of rewriting both sides which we will practice shortly. Don’t be tempted to rewrite both sides this time, that is not the goal. We’ll practice that in the next group of examples. Example: Verify that (cot x)(sin x)(sec x) = 1. Hint: Put in terms of sin & cos Example: Verify that cot2 θ(tan2 + 1) = csc2 θ is an identity. Hint: Use Pythagorean id & rewrite in terms of sin & cos Example: Verify that tan2 s sec2 s = (1 + cos s)(1 – cos s) Hint: Focus on left to restate sec & Pythagorean Id to restate tan. Example: Verify that sec s + tan s sin s = csc s sec s – tan s is an identity Hint: Focus on right. Use conjugate with fundamental theorem of fractions. Simplify & use another Pythagorean ID Now let’s practice using a different technique. Instead of focusing on just one side we’ll manipulate both sides. We’ll start with a second look at the third example above. Example: Verify that tan2 s sec2 s = (1 + cos s)(1 – cos s) Hint: Rewrite left using sin & cos. Rewrite right by mult. out & using Pythagorean Id. Example: Verify that cot θ – csc θ cot θ + csc θ = 1 – 2 cos θ + cos2 θ -sin2 θ Hint: Right: Factor, Pythagorean Id, Factor Left: Since csc & cot have sin in common, so mult by sin & simplify Now, let’s look at an application of identity verification as used in circuit design. Example: Energy in a circuit is described by E(t) = C(t) + L(t). For a certain FM radio station if the energy in the inductor is L(t) = 5 cos2 (62 x 107)t and the energy stored in the capacitor at time t is C(t) = 5 sin2 (62 x 107)t. Find the total energy in the circuit at time t. §9.2 Sum and Difference Identities I debated whether or not to cover the derivation of these formulas in class. I have decided against it due to time constraints. Suffice it to say that the proof comes from 1) 2 points on a circle with terminal points described by the sine and cosine 2) The distance formula 3) Some algebra I will leave it to a diligent student to investigate further on p. 641-642 of Ed. 5 & p. 643-644. Sine of a Sum/Difference Cosine of a Sum/Difference sin (A + B) = sinA cosB + cos A sin B cos (A + B) = cosA cosB – sin A sin B sin (A – B) = sin A cosB – cos A sin B cos (A – B) = cosA cosB + sin A sin B Tangent of a Sum/Difference tan (A + B) = tan A + tan B 1 – tan A tan B tan (A – B) = tan A – tan B 1 + tan A tan B There are multiple tasks that we will need to perform using these sum/difference identities. Some of the tasks may have characteristics from our first section. 1) We may be required to rewrite an angle as a sum of 2 angles for which we know exact trig values. 13π Ex. 15º = 45º – 30º & /12 = 9π/12 + 4π/12 = 3π/4 + π/3 Note: Here are some equivalencies that you might find helpful in doing these problems. 2π 2) 3) 4) 5) 6) 3π 4π /12 = π/6 = 30º /12 = π/4 = 45º /12 = π/3 = 60º We could be called to remember reference angles Ex. See the second example above We could be required to use even/odd identities Ex. sin (-15º) = - sin 15º We could use the identities in reverse with or without skills in 1-3 above Ex. tan 100º – tan 70º = tan (100º – 70º) 1 + tan 100ºtan 70º We might use the identities in general resulting in the use of exact trig values simplifying down to a single trig function in general Ex. Prove sin (θ – 270º) = cos θ We may use a special formula that comes from substitution in the last case to re-express a sum so that it is easier to graph as a translation of sine *See below for the formula & substitution Example: Find the exact value for a) cos (-75°) b) cos (17π/12) c) tan (13π/12) f) d) sin (-15°) Example: e) cos173ºcos83º + sin 173º sin83º tan100º – tan 30º 1 + tan 100º tan 70º Prove that cos (180 + θ) = -cos θ using the sum/difference identities. Some tools your book doesn’t give you that might be useful: The last example is a part of a group of formulas known as the reduction formulas. Reduction Formulas cos (90 + θ) = -sin θ cos (180 – θ) = -cos θ cos (180 + θ) = -cos θ cos (270 – θ) = -sin θ sin (180 + θ) = -sin θ tan (270 – θ) = cot θ Example: Prove each of the following a) sin (θ – 270º) = cos θ b) Your book makes a point to give you the following theorem. Sums of Sines and Cosines If A & B ∈ Real #’s then A sin x + B cos x = k sin (x + φ) Where k = √A2 + B2 and φ satisfies cos φ = A & sin φ = B k k tan (θ + 3π) = tan θ Example: Prove that √3/2 cos θ + -1/2 sin θ = sin (120º + θ) Example: Use what you’ve just learned to rewrite & then graph f(x) = sin x + cos x §9.3 Further Identities Double-Angle, Half-Angle & Product-Sum Formulas Double Angle Identities Half Angle Identities cos 2A = cos2 A – sin2 A √ √ √ cos 2A = 1 – 2 sin2 A 2 cos 2A = 2 cos A – 1 sin 2A = 2sin A cosA tan 2A = 2tanA 1 – tan2 A cos A = ± 2 1 + cos A 2 sin A = ± 2 1 – cos A 2 tan A = ± 2 1 – cos A 1 + cos A tan A = 2 tan A = 2 sin A 1 + cos A 1 – cos A sin A * + or – depends on quadrant of A/2 Sum to Product Product to Sum 1 cosA cos B = /2 [cos (A + B) + cos (A – B)] 1 sinA sin B = /2 [cos (A – B) – cos (A + B)] sin A cos B = 1/2 [sin (A + B) + sin (A – B)] cosA sin B = 1/2 [sin (A + B) – sin (A – B)] ( ( ( ( ) ) ) ) ( ( ( ( cos A – B 2 sin A – sin B = 2cos A + B 2 sin A – B 2 cos A + cos B = 2cos A + B 2 cos A – B 2 cos A – cos B = -2sin A + B 2 sin A – B 2 Formulas to Lower Powers sin2 A = 1 – cos 2A 2 ) ) ) ) sin A + sin B = 2sin A + B 2 cos2 A = 1 + cos 2A 2 tan2 A = 1 – cos 2A 1 + cos 2A *Note: These are just the double angle formulas from above manipulated. Some books don’t even separate them out. Hornsby, Lial and Rockswold don’t. One of the ways that we will see the DOUBLE ANGLE & HALF ANGLE formulas used is to use our basic knowledge of the 45/45/90 and 30/60/90 triangles to angles that are half or twice those values. Example: a) Practice your skills with the double angle formula by rewriting the angle to see it as twice a known angle. You probably already know all the exact values anyway, but that is not the point. cos 90º b) cos 120º c) cos π/3 d) sin π/2 e) tan -60º Example: Find the exact value of sin 22.5º by using your skill with the 45/45/90 triangle. Example: Given that sin θ = 8/17 and cos θ < 0, find sin 2θ, cos 2θ, and tan 2θ Example: Given cos x = -3/7 with π < x < 3π/2 find sin (x/2), cos (x/2), tan (x/2) Example: Use the double angle formula to find all the values of the 6 trig f(n) of θ given that cos 2θ = -12/13 and 180º < θ < 270º Example: Using a half-angle formula rewrite: √ 1 – cos 8x 2 Note: This is very simple. A = argument of the cosine so see the 1/2 angle formula and use it with A/2. Example: Verify that the following is an identity: cos4 x – sin4 x = cos 2x Hint: You’ll have to use difference of squares & an identity as well as a double angle formula. Next we can do some problems that come from the product to sum or sum to product. The product to sum identities are simple to find by adding the sum and difference formulas from our last section. Example: Write 6 sin 40º sin 15º as a sum/difference of functions. Example: Write cos 3θ + cos 7θ as a product of two functions. Here’s one that uses a few of the formulas together. Example: Write cos 3x in terms of cos x. Hint: Use a sum to rewrite 3x and then manipulate that using a formula, finally wrapping up with some double angle formulas and some simplification. §9.4 The Inverse Circular F(n) First let’s review relations, functions and one-to-one functions. A relation is any set of ordered pairs. A relation can be shown as a set, a graph or a function (equation). a) b) c) {(2,5), (2,6), (2,7)} y = √x, {x| x ≥ 0, x∈ℜ} 5 -15 x 15 -5 A function is a relation which satisfies the condition that for each of its independent values (x-values; domain values) there is only 1 dependent value (y-value; f(x) value; range value). From above, only satisfies this requirement b) y = √x, {x| x ≥ 0, x∈ℜ} Note: We can check to see if a relation is a f(n) by seeing if any of the x’s are repeated & go to different y’s; on a graph a relation must pass vertical line test and if it’s a f(n)/equation then you can think about it in terms of the domain and range or in terms of its graph. A one-to-one function is a function that satisfies the condition that each element in its range is used only once (has a unique x-value; domain value). From above, only satisfies this requirement b) y = √x, {x| x ≥ 0, x∈ℜ} Note: We can check to see if a function is 1:1 by seeing if any of the y’s are repeated & go to different x’s; on a graph a 1:1 f(n) must pass horizontal line test and if it’s a f(n)/equation then you can think about it in terms of the domain and range or in terms of its graph. If you remember from your study of Algebra, we care about one-to-one functions because they have an inverse. The inverse of a function, written f-1(x), is the function for which the domain and range of the original function f(x) have been reversed. The composite of an inverse and the original function is always equal to x. Inverse of f(x) f -1(x) = {(y, x) | (x, y) ∈ f(x)} Note: f -1(x) is not the reciprocal of f(x) but the notation used for an inverse function!! The same will be true with our trigonometric functions. Facts About Inverse Functions 1) Function must be 1:1 for an inverse to exist; we will sometimes restrict the domain of the original function, so this is true, but only if the range is not effected. Note: You will see this with the trig functions because they are not 1:1 without the restriction on the domain. 2) 3) 4) 5) (x, y) for f(x) is (y, x) for f -1(x) f(x) and f -1(x) are reflections across the y = x line The composite of f(x) with f -1(x) or f -1(x) with f(x) is the same; it is x The inverse of a function can be found by changing the x and y and solving for y and then replacing y with f -1(x). Note: This isn’t that important for our needs here. What you should take away from the above list is: 1) We make the trig functions one-to-one by restricting their domains 2) (x, y) is (y, x) for the inverse function means that if we have the value of the trig function, the y, the inverse will find the angle that gives that value, the x 3) When we graph the inverse functions we will see their relationship to the graphs of the corresponding trig function as a reflection across y = x. 4) sin -1 (sin x) is x Note: You might use this one in your Calculus class. The Inverse Sine – Also called the ArcSin If f(x) = sin x on D: [-π/2, π/2] & R: [-1, 1] Notice: Restriction to the original domain is to QI & QIV of unit circle! then f -1(x) = sin-1 x or f -1(x) = arcsin x Note: Scan from p. 551 Stewart ed 5 on D: [-1, 1] & R: [-π/2, π/2] Note: Scan from p. 551 Stewart ed 5 Finding y = arcsin x 1) Think of arcsin as finding the x-value (the radian measure or degree measure of the angle) of sin that will give the value of the argument. Rewrite y = arcsin x to sin y = x where y = ? if it helps. 2) Think of your triangles! What ∠ are you seeing the opposite over hypotenuse for? Example: Find the exact value for y without a calculator. Don’t forget to check domains! Use radian measure to give the angle. a) y = arcsin √3/2 b) y = sin -1 (-1/2) c) y = sin -1 √2 However, it is not always possible to use our prior experience with special triangles to find the inverse. Sometimes we will need to use the calculator to find the inverse. Example: Find the value of the following in radians rounded to 5 decimals *Precalculus, 6th ed., Stewart, Redlin & Watson p. 467 #10 y = sin -1 (1/3) The Inverse Cosine – Also called the ArcCos If f(x) = cos x on D: [0, π] & R: [-1, 1] Notice: Restriction to the original domain is to QI & QII of unit circle! then f -1(x) = cos-1 x or f -1(x) = arccos x on Note: Scan from p. 553 Stewart ed 5 D: [-1, 1] & R: [0, π] Note: Scan from p. 553 Stewart ed 5 Example: Find the exact value of y for each of the following in radians. a) y = arccos 0 b) y = cos -1 (1/2) Example: Find the value of the following in radians rounded to 5 decimals *Precalculus, 6th ed., Stewart, Redlin & Watson p. 467 #8 y = cos-1 (-0.75) The Inverse Tangent – Also called the ArcTan If f(x) = tan x on D: [-π/2, π/2] & R: [-∞, ∞] Notice: Restriction to the original domain is to QI & QIV of unit circle! then f -1(x) = tan-1 x or f -1(x) = arctan x Note: Scan from p. 555 Stewart ed 5 on D: [-∞, ∞] & R: [-π/2, π/2] Note: Scan from p. 555 Stewart ed 5 Notice: The inverse tangent has horizontal asymptotes at ±π/2. It might be interesting to note that the inverse tangent is also an odd function (recall that tan (-x) = -tan x and also the arctan (-x) = - arctan (x)) just as the tangent is and that both the x & y intercepts are zero as well as both functions being increasing functions. Example: Find the exact value for y, in radians without a calculator. y = arctan √3 I am not going to spend a great deal of time talking about the other 3 inverse trig functions in class or quizzing/testing your graphing skills for these trig functions. I do expect you to know their domains and ranges and how to find exact and approximate values for these functions. Summary of Sec-1, Csc-1 and Cot-1 Inverse Function Domain y = cot-1 x (-∞, ∞) -1 y = sec x (-∞,-1]∪[1,∞) -1 y = csc x (-∞,-1]∪[1,∞) Example: Interval (0, π) [0, π], y≠π/2 [-π/2, π/2], y≠0 Quadrant on Unit Circle I & II I & II I & IV Find the exact value for y, in radians without a calculator. y = arccsc -√2 Before we do any examples that require the use of our calculator to find the inverse of these functions, it is important that you recall your reciprocal identities! It is by using these reciprocal identities first that you can find the inverse values on your calculator. Let me show you an example first. Example: Find the approximate value in radians rounded to the nearest 5 decimals. y = sec-1 (-4) STEP 1: Rewrite as secant sec y = -4 STEP 2: Rewrite both sides by taking the reciprocal 1/sec y = -1/4 STEP 3: Use reciprocal identity to rewrite cos y = -1/4 STEP 4: Find the inverse of both sides to solve for y cos-1(cos y) = cos-1 (-1/4) STEP 5: Use your calculator to find y = cos-1 (-1/4) Example: Find the approximate value in radians rounded to the nearest 5 decimals. y = arccot (-0.2528) We can also “take a page” from the skill-set that we have just been using and use it to solve a little more difficult problems – those that deal with the composite of trig and inverse trig functions. In these type of problems we will work from the inside out. We don’t actually ever get a degree measure from these problems, because the final answer is actually the ratio of sides. 1) We will use the inside function to put together a triangle 2) Then we use the Pythagorean Theorem to find the missing side 3) Finally, we use the definitions of the trig functions based upon opposite, adjacent and hypotenuse lengths to answer the problem. Example: Evaluate each of the following without a calculator a) cos (sin-1 2/3) b) sec (cot-1 (-15/8)) Note that I have only given examples that deal with radians, but there are also problems that ask for the answers in degrees. Practice both and remember when it comes to applications that involve arc length the use of radians is necessary. See #42 p. 468 of edition 6 for example. Let’s do an application problem that will remind us of our previous instruction on angles of elevation. Example: Find the angle of elevation of the sun. *Precalculus, 6th ed., Stewart, Redlin & Watson p. 468 #38 96 ft. θ 120 ft. §9.5 Solving Basic Trig Equations Recall conditional linear equations in Algebra had 1 solution. We are going to study conditional trig equations first. This means that some values will solve them and some will not. Method 1: Linear Methods 1) Employ algebra to isolate the trig function 2) Use trig identities as needed 3) Use definitions as needed 4) When using the inverse to solve, make sure that you realize that the answer you get comes from the range of the inverse function & you may need to use that as a reference angle(point) to find all possible values for original equation. Recall: sin(sin-1 x) on [-1, 1] sin-1 (sin x) on [-π/2, π/2] -1 cos(cos x) on [-1, 1] -1 cos (cos x) on [0, π] tan(tan -1 x) on (-∞, ∞) -1 tan (tan x) on (-π/2, π/2) 5) Also realize that when finding all possible solutions that each of the functions repeats with a certain period and once the solutions are found in the fundamental period they can then be expanded to encompass all periods. Recall: sine & cosine’s period is 2π ∴ 2 values w/in fundamental cycle + 2kπ where k is an integer tangent’s period is π ∴ 2 values w/in fundamental cycle + kπ where k is an integer Example: Solve 3 tan θ – √3 = 0 Method 2: Factoring & Quadratic Methods 1) Get f(x) = 0 using algebra 2) Factor & use zero factor property 3) Use Method 1’s process if needed Example: Solve tan x sin x + sin x = 0 on [0, 360º) Example: Solve cos θ cot θ = -cos θ on [0, 360º) Sometimes we will have extraneous roots that we will need to catch as we did when solving rational equations in algebra. Method 3: Using Trig Identities 1) See a relationship that exists between 2 functions 2) Square sides in order to bring out an identity – be careful to check solutions for extraneous roots any time you do this 3) Use an identity to rewrite in terms of one function 4) Solve as in the last section Example: Solve cot x – √3 = csc x on [0, 2π) Note: csc and cot are related via a Pythagorean Identity. You can create a problem by squaring to get there, so do check extraneous roots. Example: Solve cos 2x = sin x Note: Did you get 3 solutions? Don’t forget about reference angles. on [0, 2π) Example: Solve 2 cos2 θ – 2 sin2 θ + 1 = 0 on [0, 360º) Note: 1st because the period is cut in half, the function will oscillate twice the number of times on [0, 360º), so we need to find the answers on [0, 720º) and furthermore we end up solving for 2A, we need to make sure we answer for A by dividing all by 2! Example: Solve cos 2x + cos x = 0 §9.6 Trig Equations and Inequalities When solving a trig inequality, using a graph is the easiest method. Recall that solving a trig equation graphically requires looking for the x-intercepts. Let’s take a look at the graphic solution for the second to the last example: Example: 1) 2) 3) 2 cos2 θ – 2 sin2 θ + 1 = 0 Rewrite using the double angle formula for cos 2 θ Investigate the domain and range of the function a) Domain [0, 2π) b) Range [-1, 3] c) Scale π/6 The on [0, 2π)