Chapter 06: Analytic Trigonometry

Chapter 06: Analytic Trigonometry 6.1: Inverse Trigonometric Functions The Problem As you recall from our earlier work, a function can only have an inverse function if it is oneto-one. Are any of our trigonometric functions one-to-one? Alas, no which is a problem! There are times when we would really like to be able to find an angle, given the value of, say, the sine ratio The Solution There is a way we even did it a few times before to force a function to have an inverse: restrict the domain of the function. Consider the sine function: Can you see a domain restriction (set of x-values) that would make the function one-to-one? (there are several) Restrictions Here are the standard restrictions: Function Restricted Domain Sine π π, Cosine [ ] 0,π π π Tangent, Remember that the domain and range switch places once we actually find the inverse function. Also, note that these inverse functions are only defined for radian angle measures! Thus, we can summarize the basic inverse trigonometric functions: Function Domain Range π π y = sin x [ 1,1] y, y cos x 1,1 y 0, π y = x [ ] [ ] = tan x R y, π π HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE 1 OF 1

Using Your Calculator These basic inverse trigonometric functions are built into your calculator, so no special techniques are needed to use them. [1.] Find sin. π 1 Since sin = 1 1, sin π =. There are an infinite number of angles whose sine value is 6 6 one half, but this angle is the only one that is in the range of the inverse sine function. 1 1 [.] Find cos. π 1 Since cos =, 4 1 1 1 π cos =. The same comment above applies here, too. 4 6.: More Inverse Trigonometric Functions Mixing Regular and Inverse Functions Note that the inverse functions take in a number and spit out an angle; thus, an expression like cos sin is really just cosine of some angle. One way to deal with expressions like this is to draw a picture of that mystery angle, and then find the trig function value of that angle. The situation is more difficult if you have the inverse function on the outside like, say sin cos π. In cases like this, the inner trig function must produce a value which is one of ( ) your memorized trig values if you re to have any hope of working out the value without a calculator. 7π Of course, situations like sin sin 6 are usually easier but be careful! 7π 7π sin sin 6! Remember the restrictions on the inverse functions 6 The Other Inverse Functions There are three other trigonometric functions whose inverse we haven t looked at yet let s have at it, then. Function Domain Range π π y = sec x (, ] [1, ) y [0, ) (, π ] HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE OF 1

y y x π π y [,0) (0, ] y 0, π = csc (, 1] [1, ) = cot x R There is some disagreement about these restrictions in fact, there are many who don t even bother to use these inverse functions at all, instead relying on just the basic three. Using Your Calculator The domain restrictions are one problem; calculator issues are the other issue. Very few, if any calculators have an inverse secant function thus, the problem must be reworked so that one of the basic trig functions is being used, and then the problem is solved using the basic inverse trig functions. [.] Find Let s just say that cos sin. sin = θ. I can draw a triangle that uses that angle: 1 θ Note that the other leg must have length 5. So, I can now find cos( θ ) it is 5. Thus, 5 cos sin =. I know that the answer must be positive since inverse sine of a positive number must return an angle in the first quadrant, and cosine of that type of angle is positive. [4.] Find sec. 1 Let s say sec = a π ; thus, sec( a) = cos( a) = a = cos =. This is 6 how you d need to approach the problem with a calculator, also. 6.: Trigonometric Identities The Process We ve encountered proofs before! The main things to remember are [1] you can t work both sides, [] pick the more complicated side to work on, and [] don t panic! HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE OF 1

Working these proofs requires that you see work a lot on your own, and that you see and recognize patterns in the proofs. There are plenty of problems to practice in the textbook! [5.] Show that ( 1 sin ( θ ))( 1 sin ( θ )) cos ( θ ) + =. Well I m certainly going to work the left side of that! 1 sin θ 1+ sin θ = 1 sin θ, which looks a lot like the Pythagorean identity: ( )( ) ( θ ) ( θ ) ( θ ) ( θ ) sin + cos = 1 cos = 1 sin. Thus, I ve done it. That was easy! [6.] Show that tan cot sec csc + =. Again, I ll work the left side of that. sin cos sin sin cos cos x tan + cot = + = + cos x sin x sin x cos x cos x sin x + sin x cos x sin x cos x 1 = + = = sin x cos x sin x cos x sin x cos x sin x cos x = 1 1 sec csc sin =. cos 1+ sin α cos α [7.] Show that =. cos α 1 sin α Hmmm neither side looks more complicated but I like the denominator on the right less, so I ll work that side. Note the trick! cos ( α ) 1+ sin ( α ) ( cos( α ))( 1+ sin ( α )) ( cos( α ))( 1+ sin ( α )) 1+ sin ( α ) = = =. 1 sin α 1+ sin α 1 sin α cos α cos α 6.4: Sum and Difference Formulas Deriving the Formulas Let s see where these come from! I ll use the unit circle approach. Cosine Let s draw a picture of the angle α β. HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE 4 OF 1

The red line and coordinates are there for a reason! Now, the angle α β is not in standard position so let s rotate those two points on the unit circle until the angle is in standard position. Note that the red line must have the same length in this picture as it did in the first and that s the key. Distance formula, anyone? Distance from the top picture first: ( cos( α ) cos( β )) + sin ( α ) sin ( β ) ( α ) ( α ) ( β ) + ( β ) + ( α ) ( α ) ( β ) + ( β ) cos cos cos cos sin sin sin sin ( α ) ( β ) ( α ) ( β ) cos cos sin sin Now let s work on the distance in the second picture: ( ( α β ) ) ( α β ) cos 1 + sin 0 ( α β ) ( α β ) + + ( α β ) cos cos 1 sin cos( α β ) Let s finally set those equal to each other: cos( α ) cos( β ) sin ( α ) sin ( β ) = cos( α β ) cos( α ) cos( β ) sin ( α ) sin ( β ) = cos( α β ) cos( α ) cos( β ) sin ( α ) sin ( β ) = cos( α β ) cos( α ) cos( β ) + sin ( α ) sin ( β ) = cos( α β ) and there it is. Replace β with β and you ll get the formula for the sum of two angles. HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE 5 OF 1

In summary: cos α ± β = cos α cos β sin α sin β Sine Here are two important identities that will establish the formulas for sine: π sin = cos x ; cos sin π x = x So if we want to find the formula for the sine of a sum, we ll do this: π π sin ( α + β ) = cos ( α + β ) = cos α β Now apply the cosine formula that we just developed! π π π cos α β = cos α cos( β ) + sin α sin ( β ) Finally, use those two identities again: π π cos α cos ( β ) + sin α sin ( β ) = sin ( α ) cos( β ) + cos ( α ) sin ( β ) Replace β with β and you ll get the formula for the difference of two angles. In summary: sin α ± β = sin α cos β ± cos α sin β Tangent Tangent is defined as sine over cosine use our already developed formulas to create one for tangents! sin ( α + β ) sin ( α ) cos ( β ) + cos( α ) sin ( β ) tan ( α + β ) = = cos α + β cos α cos β sin α sin β Looks nasty because there s one more step: divide both the numerator and denominator by cos α cos β. + + sin α cos β cos α sin β sin α sin β + cos α cos β cos α cos β tan α tan β = = cos α cos β sin α sin β sin α sin β 1 tan α tan β 1 cos α cos β cos α cos β Replace β with β and you ll get the formula for the difference of two angles. In summary: tan ( α ) ± tan ( β ) tan ( α ± β ) = 1 tan α tan β π [8.] Evaluate cos 1. HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE 6 OF 1

π 4π π π π Note that = =. Thus, 1 1 1 4 π π π π π π π cos = cos = cos cos + sin sin = 1 4 4 4 1 1 1 1+ + =. 1 [9.] Evaluate sin cos + cos 5. This requires a bit from this chapter and a bit from two chapters ago! 1 1 1 sin cos + cos = sin cos cos cos + cos cos sin cos 5 5 5 1 Draw some triangles to find sin cos and sin cos 5. The result is 8 1 1 8 + 1 + =. 5 5 15 π tan [10.] Show that tan x =. 4 tan + 1 Naturally, I ll work the left side: π tan tan π 4 tan 1 tan 1 tan x = = =. QED. 4 π 1+ tan tan + 1 1+ tan x tan 4 6.5: Double and Half Angle Formulas The Double Angle Formulas Simply replace β with α in each formula! sin = sin cos cos = cos sin tan tan = 1 tan The Pythagorean Identity allows you to make some changes to the cosine formula, resulting in cos x cos x 1 cos x = 1 sin x. either = or The Power Reducing Formulas Take those last two formulas and solve them for the squared term. These identities are really useful in Calculus! HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE 7 OF 1

1+ cos cos = 1 cos sin = If you put those two together, you get a power reducing identity for tangent: 1 cos tan = 1 + cos x The Half Angle Formulas Take those identities, square root them and then replace x with x to obtain the half angle identities! x 1+ cos cos = ± x 1 cos sin = ± x 1 cos x tan = ± 1 + cos Rearranging things a bit can make that last tangent identity look like this: x 1 cos sin tan = = sin x 1 + cos x [11.] Find sin ( x ) if sin π x = and < x < π. 4 Draw a triangle to see that this makes cos 7 7 sin = sin cos = 4 = 4. 8 x = 7 4. Thus, 1 sin ( θ ) cot ( θ ) [1.] Show that = cos ( θ ) cot ( θ ) + 1. I m going to work the right side of this and I ll start by multiplying the top and bottom by sin ( θ ). HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE 8 OF 1

θ θ θ θ θ θ θ θ θ + θ cot 1 sin cos sin cos sin cos sin cos sin = = cot θ + 1 sin θ cos θ + sin θ cos θ sin θ cos θ sin θ 1 sin θ cos θ 1 sin θ = =. QED. θ θ θ cos sin cos [1.] Evaluate 1 tan cos. 1 cos cos 1 1 1 1 1 tan cos = = = = 5 1 5 1+ cos cos + 6.6: Product/Sum Formulas Turning Products into Sums You know that cos( α β ) cos( α ) cos( β ) sin ( α ) sin ( β ) cos( α β ) cos( α ) cos( β ) sin ( α ) sin ( β ) cos( α β ) cos( α β ) sin ( α ) sin ( β ) = + and that + =. Subtract these two and you get + = which can be rearranged to sin ( α ) sin ( β ) = 1 cos ( α β ) cos( α + β ). A little algebra can be employed to make two other versions of this! cos ( α ) cos( β ) = 1 cos ( α β ) + cos ( α + β ) sin ( α ) cos( β ) = 1 sin ( α + β ) + sin ( α β ) Turning Sums into Products α + β α β Rearranging those formulas, and replacing the angles with and results in formulas that turn addition into multiplication. α ± β α β sin ( α ) ± sin ( β ) = sin cos α + β α β cos ( α ) + cos( β ) = cos cos α + β α β cos ( α ) cos( β ) = sin sin HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE 9 OF 1

[14.] Express as a sum of sines: sin cos( x ). sin cos = 1 sin ( x + x) + sin ( x x) = 1 sin ( 4x) + sin ( x) = 1 sin ( 4 x) sin ( x) [15.]Express as a sum of cosines: sin sin ( x ). sin sin = 1 cos ( x + x) cos ( x x) = 1 cos ( 4x) cos( x) = 1 cos ( 4 x) cos ( x) [16.] Express as a product of cosines: cos cos( 5x) +. x + 5x x 5x 4x 6x cos ( x) + cos( 5x) = cos cos = cos cos cos x cos x = cos x cos x 6.7: Solving Trigonometric Equations Finding the General Solution For many equation involving trigonometric functions, there are an infinite number of solutions in these cases, we usually find all solutions over on period of the function, and then add a little notation at the end that says add or subtract any number of periods you want. If the problem says to give the general solution, or if the problem says to find all solutions, then you will use this method (illustrated below). Solving on an Interval One of the most common instructions is to solve over some given interval. This is typically much easier! The exception is if the angle used in the trig function is a multiple ( x or x, for example). [17.] Give the general solution to tan ( x ) =. First, the period of tangent is π so I ll find all solutions in the interval [ 0.,π ]: π tan = x =. Now, to that I need to add any (integer) number of periods here s how π that looks: x = + kπ, k Z. HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE 10 OF 1

[18.] Solve sin cos = for x [0, π ). Since an interval is given, I just need to find all of the solutions on the given interval. π 5π sin = cos tan = 1 x, 4 4. [19.] Find all solutions to sin = sin. The period of sin ( x ) is π, and the period of initial solutions. sin cos 1 = 0. sin = 0 x { 0, π} ; sin x is π I ll use the larger one for my sin = sin sin cos = sin sin cos sin = 0 1 π 5π cos 1 = 0 cos = x,. Thus, my final solution will be π 5π x 0,, π, + kπ, k Z. [0.] Solve tan x = for x [0, π ). This is where you must be careful since x [0, π ), x [0, π )! π 5π 7π 11π tan = tan = ± x,,, 6 6 6 6. That makes π 5π 7π 11π x,,, 1 1 1 1. 6.8: Solving More Trigonometric Equations Quadratic Trigonometric Equations x x Do you remember when we saw that e e + 1 was really a quadratic expression? Well, that s about to happen again this time with trig functions. Other Methods in other words, this section will require solving methods other that just quadratic [1.] Solve cos + cos = 1 for x [0, π ). cos + cos = 1 cos + cos 1 = 0. Let u cos =, making the equation 1 u + u 1 = 0 ( u 1)( u + 1) = 0, so either u 1 = 0 u = or u + 1 = 0 u = 1. Since 1 π 5π u = cos, that gives us cos = x, or cos = x = π. π 5π The solutions are x, π,. HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE 11 OF 1

[.] Solve cos = sin for x [0, π ). cos = sin cos sin = 0 cos = 0. Since x [0, π ), x [0,4 π )! π π 5π 7π cos = 0 x,,, which means that π π 5π 7π x,,, 4 4 4 4. (I can think of one other way to do this ) HOLLOMAN S PRECAL HONORS CHAPTER 06 NOTES, PAGE 1 OF 1