Trigonometry Review Workshop 1

Transcription

1 Trigonometr Review Workshop Definitions: Let P(,) be an point (not the origin) on the terminal side of an angle with measure θ and let r be the distance from the origin to P. Then the si trig functions are defined as follows: r θ sinθ = opposite hpotenuse = r cscθ = r = sinθ cosθ = adjacent hpotenuse = r secθ = r = cosθ tanθ = opposite adjacent = = sinθ cosθ cotθ = = tanθ = cosθ sinθ Measuring Angles There are 60 o in a circle and π radians. The are marked off on the circle below: π/, 0ο π/, 90ο π/, 60ο π/4, 5ο π/4, 45 ο 5π/6, 50 ο 0, π/6 ο π, 80 π, 0ο ο 7π/6, 0 ο π/6, 0 ο 5π/4, 5 7π/4, 5 ο 4π/, 40ο ο π/, 70 5π/, 00 ο To convert from degrees to radians (and vice versa) we use the fact that 80 o = π radians.

2 Eample: Convert 6 o to radians. π Using the ratio 80 o we find that 6o = 6 o π 80 o = π 0 = π 5 radians Reference Angles The angle formed b the terminal side of the angle and the -ais is called the reference angle. The si trig functions of an angle are determined b it's reference angle and the quadrant it is in. R R R R = 80 o - θ R = θ 80 ο R = 60 o - θ The signs of the three major trig functions are determined b the quadrant as follows: Quadrant I: All are positive Quadrant II: Sine is positive Quadrant III: Tangent is positive Quadrant IV: Cosine is positive This can be remembered b the pneumonic device All Students Take Calculus. Since cscθ =, sine and cosecant alwas have the same sign. The is true of Cosine, Secant, and Tangent, sinθ Cotangent as well. Special Triangles: We can use the ratios of some common triangles, along with the definitions of the trig. functions to determine the trig functions of some special angles. Using the two triangles below we see that:

3 sin 45 o = = sin 0 o = cos 45 o = = cos 0 o = tan 45 o = = tan 0o = = and similarl for 60 o. This is left as an eercise. Practice Eercises #. Find the si trig functions of 60 o.. Find the si trig functions for each of the following angles as shown: a. b. 5 θ 5 θ. Find the si trig functions of each of the following angles: a. 5 o b. 5π 6 4. Given that sinθ = and θ is in quadrant IV, find the cosine and tangent of θ. 5. Convert 4 o into radians. 6. Convert 7π 8 radians into degrees.

4 Trigonometr Review Workshop Last time we saw the definitions of the si trig. functions and evaluated them for some special angles. There are a few more "nice" angles which we will discuss here. Another wa to define the functions sinθ and cosθ is as follows. If we look at the unit circle below, (0,) (,) (-,0) θ (,0) (0,-) and let (,) be an point on the circle, then and cosθ=, and sinθ =. To see this simpl eamine the triangle above and use the previous definitions of these functions. Thus to find the sines and cosines of angles we onl need to find the and components of the point on the unit circle that corresponds to that angle. Hence we see that: cos0 o = cos 90 o = 0 cos 80 o = cos 70 o = 0 sin 0 o = 0 sin 90 o = sin 80 o = 0 sin 70 o = Since we know the sines and cosines for these angles we can find values for the other si trig functions: Fill in these for 0 o and 90 o below: tan 0 o = cot 0 o = sec 0 o = csc 0 o = tan 90 o = cot 90 o = sec 90 o = csc 90 o = At this point ou should be able to calculate the trig functions of all angles with reference angles of 0 o, 45 o, 60 o, as well as all multiples of 90 o. For other angle it will be necessar to use a calculator. Don't forget to put our calculator in the appropriate mode (either degrees or radians) depending on which ou are using.

5 Trigonometric Identities: The following are some of the major trig identities which ou will need to know. This list is NOT a complete list of all trig identities, simpl the important ones. Pthagorean Identities: sin θ + cos θ = + cot θ = csc θ (Note that the last two can be derived from the first b tan θ + = sec θ dividing b either sin θ or cos θ) Addition formulas: sin(a±b) = sinacosb ± cosasinb cos(a±b) = cosacosb + sinasinb (Note that for the cosine addition formula the +/- are reversed) If we let A=B in the above formulas and choose the "+" we get the Double Angle Identities: sin(a) = sinacosa cos(a) = cos A - sin A In addition we have the following laws which appl to triangles: A Law of Cosines: a = b + c -bc cosa c b Law of Sines: B sina = sinb = sinc a C a b c Practice Eercises #:. Find all sides and angles in the following triangles: a. b o a θ 4 0 o a b 0 o b c. If sin(a) = / and A is in quadrant II, find: (a) sin(a); (b) cos(a); (c) tan(a). Solve for all values of 0 π that satisf: (a) sin = 4. Verif the identit sin θ + cos θ = for θ=0 ο. 5. Find the si trig functions of 80 o (b) cos sin = sin

6 Trigonometr Review Workshop This time we will look at the graphs of = sin() and = cos(), as well as some variations. Recall last time we defined the functions sinθ and cosθ is as follows. If we look at the unit circle below, (0,) (,) (-,0) θ (,0) (0,-) and let (,) be an point on the circle, then and cosθ=, and sinθ =. Notice that as we move around the circle = sin(), which measures the value of the point starts at 0 and increases to at = π/. Cosine however, which measures the value of the point starts at then decreases to 0 at = π/. The are both described in the tables below: = sin() = cos() 0 0 π/ 0 π 0 - π/ - 0 π 0 Thus we have the following graphs of = sin(), and = cos () respectivel: = sin() = cos () Now what happens if we change things a bit? For eample, let's look at the graph of = sin(). If we complete the chart as we did above we have:

7 and the graph: = sin() = sin() π/ π 0 0 π/ - - π 0 0 The graph has effectivel been stretched verticall. This stretch changes the aplitude of the graph. The Amplitude of = Asin() is A. Now let's consider that graph of = sin(). Notice that this differs from the above function in that the is inside that sine (not outside). = sin() π/4 π/ π/ π 0 π/4 π/ - π π 0 Notice that the graph has been "shrunk" horizontall b /. This changes the period of the function (or the distance before the function begins repeating itself).

8 In general the period of = sin(b) is π / B. Consider the graph of = sin()+. Notice that we are simpl adding one to the -value for each -value. This has the effect of shifting the graph verticall b one: As our last eample consider the function = cos( - π/). Since we are subtracting from the -value (not the -value) this will shift the graph horizontall. Although it might seem reasonable to epect a shift to the left, the graph will be shifted to the right. The easiest wa to determine the shift is to set the angle of the funtion to zero. In this case that means setting - π/ = 0 and so = π/. Thus the shift is to the right b a positive π/, and we have the graph: Practice Ecercises #:. Find the period of each of the following: (a) = sin(5) (b) = cos (/) (c) = sin (π). Graph each of the following on a seperate ais. (a) = sin (4); (b) = cos(π)+; (c) = sin ( + π). Find an eample of a sine function which has an amplitude of, period of 5π and passes through the point (0,4)

9 Trigonometr Review Workshop 4 In this workshop we will continue to discuss solving trigonometric equations and further identites. Eample: Solve sin = for all 0 < π. Solution: This problem is ver similar to solving the equation = and should be thought of as similar. There are two was to approach solving it. One is to take the square root of both sides. The other is to put everthing on one side and factor. The latter provides a more general wa so that is the approach that will be taken here. sin = sin - = 0 (sin +)(sin -) = 0. Hence we have sin = or sin =. Recall that the sine represents the vertical distance on the unit circle. B eamining the unit circle we see that sin = when = π/ and sin = when = π/ radians. Hence, the solutions set is {π/,π/ }. Notice that we do not include π, since it is not in the interval 0 < π. Eample: Solve cos =, where 0 < π. There are several was to approach this problem. One possible wa would be to use the double angle formula for cosine. Let's solve the problem this wa, then consider a different approach. cos = cos sin = ( sin ) sin = sin = sin = 0 sin = 0 sin = 0. Since sin = 0 at = 0 and = π, our solution set is {0, π}. There is, however, another approach to this problem. Since is the angle, let's solve for first. Notice that cos has a period of π. Hence, to an solution we obtain, we can add the period of this function and this will give us another solution. In solving cos = notice that there is onl one place where the cosine is zero and that is at 0 radians. Hence, =0, and thus =0. Adding one period to this answer gives a second answer 0+ π = π. Hence the solutions are = 0 and = π radians. Notice that if we add π again we get π + π = π, which is outside the range of answers we want. The solution set {0, π} is complete. Notice that this agrees with the answer above. Eample: Solve sin = for 0 < π. Solution: First take the square root of both sides to obtain sin = ±. That is, sin =, or sin =. The sinθ = at θ = π/ and the sinθ = at θ = π/. Hence we have = π/, and = π/, or = π/6, and = π/. Again, this is not a complete set of solutions. The period of sin is π/ and this needs to be added to each solution until the values eceed π. Hence another solution is π/6+π/ = 5π/6. Continuing this process, we obtain the complete set of solutions: { π/6, π/, 5π/6, 7π/6, π/, π/6}.

10 More Identities Trig identities pla an important part in calculus. Being able to identif the major ones, especiall the double angle formulas and the famous sin + cos = is ver important. There are, however, man other useful identities which can be proven using definitions and eisting identities. For eample, it can be shown that cos + sin =. To show this, start with one side and tr to reduce it to the other. For this eample (and in general) it is generall easier to start with the more difficult side and tr to simplif it. Start b using the double angle formula for cosine: cos + sin = (cos -sin )+sin, = cos + ( sin + sin ), = cos + sin, =. Eample: Prove the identit (sec )(cot )(csc ) = csc. Solution: Beginning with the left hand side, and using the basic definitions of the trig functions we find: Practice Ecercises #4: (sec )(cot )(csc ) = cos = sin = csc. cos sin sin. Solve each of the following trigonometric equations: (a) sin() = / (b) sec ()= (c) cos + sin = (d) tan (4) =. Verif each of the following identities: (a) (sin ) (sec ) = tan (b) (tan )(sin ) = sin (c) tan cos = sec sin (d) cos 4 sin 4 = -sin.

11 Solutions to Problem Set #:. sin 60 o = 60 o =. (a) sin θ =, sec 60o =, cos 60 o =, csc 60o =, cot 60 o =, tan 5, csc θ = 5, cos θ = 5, sec θ = 5, tan θ =, cot θ=, (b) sin θ =, csc θ = - 6, cos θ =, sec θ = 6 6 5, tan θ = - 5, cot θ= -5,. sin(5) =, csc(5) =, cos(5) =, sec(5) =, tan(5) =, cot(5) = ; sin(5π/6) = /, sec(5π/6) =, cos(5π/6) = /, sec(5π/6) = /, tan (5π/6) = /, cot(5π/6) = 4. cosθ = π 45 radians, 6. 7π 8 radians = 57.5 o Solutions to Problem Set #:, tanθ = -, 5. 4 o = () (a) a = b = 5,θ = 45 o (b) a = 4, b = 0 o, c = 4 () (a) 4 9 (b) 7 9 (c) 4 7 () (a) = π, π (b) = 0, π,π, 5π (4) sin 0 + cos 0 = ( ) + ( ) = =. (5) sin(80 o ) = 0, cos(80 o ) =, tan(80 o ) = 0, csc(80 o ) is undefined, sec(80 o ) =, cot(80 o ) is undefined. Solutions to Problem Set #:. (a) π/5, (b) 4π, (c). (a) (b) (c) π/ π π. = sin Solutions to Problem Set #4. (a) = π/, 5π/, π/, 7π/ (b) = π/, π/, 5π/, 7π/, 9π/, π/, π/, π/, 5π/, 7π/, 9π/, π/, π/. (c) = 0, π/, π (d) = π/, 5π/6, 4π/, π/6