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Unformatted text preview: Math 1205 Trigonometry Review We begin with the unit circle. The definition of a unit circle is: x 2 + y 2 = 1 where the center is (0, 0) and the radius is 1. An angle of 1 radian is an angle at the center of a circle measured in the counterclockwise direction that subtends an arc length equal to 1 radius. Notice that the angle does not change with the radius. There are approximately 6 radius lengths around the circle. That is, one complete turn around the circle is 2p 6.28 radians. Define the Sine and Cosine functions: Choose P(x, y) a point on the unit circle where the terminal side of q intersects with the circle. Then cosq = x and sin q = y . We see that the Pythagorean Identity follows directly from these definitions: x2 + y2 =1
(cosq ) 2 + (sin q ) 2 = 1 we know it as : sin 2 q + cos2 q = 1
Example 1. Determine:
sin(90) and cos(90) Example 2. Determine: sin(3p ) and cos(3p ) Recall that 90 corresponds to p radians . 2 (How many degrees do 3p radians correspond to?) We can read the answers from the graphs: p ^ sin(90) = sin ~ = y coordinate of P = 1 2 p ^ cos(90) = cos ~ = x coordinate of P = 0 2 sin(3p ) = sin(540) = y coordinate of P = 0 cos(3p ) = cos(540) = x coordinate of P = 1 Problems 1 and 2: 1. Locate the following angles on a unit circle and find their sine and cosine. 5p 5p a. b. c. 360 d. p 2 2 2. Given: cosq = 0 and sin q = 1 . Find the following: a. the smallest positive q that satisfies the given equalities. b. one other q that satisfies the given equalities. Note: q1 and q 2 should be in radians. There are six trigonometric functions. We have considered the sine and cosine functions. We can define the four remaining in terms of these functions. sin q cosq The tangent function: tanq = The cotangent function: cot q = cosq sinq 1 The cosecant function: csc q = sin q 1 The secant function: sec q = cos q We know all of the above functions will have points of discontinuity where the denominator is zero. The graphs of these functions all have vertical asymptotes at these points. We will use the definition of the sine and cosine functions on the unit circle ( r = 1) to find the sine and cosine for common reference angles. cosq = x and sin q = y We could use the sine and cosine graphs, however the unit circle is more useful for these problems. The common angles that we are interested in are:
degrees radians 0 0 30 /6 45 /4 60 /3 90 /2 180 270 3/2 360 2 Consider p = 45 4 An angle of p radians intersects the unit circle at the point, 4 1 1 ^ P = , ~. 2 2 Using the definition for sine and cosine, we have: p ^ 1 p ^ 1 sin ~ = and cos ~ = 4 4 2 2 Similarly, we find the sine and cosine of p = 30 : 6 p ^ 1 p ^ 3 sin ~ = and cos ~ = 6 2 6 2 We can complete the chart by working in the same manner to get:
degrees radians sin q cos 0 0 0 1 30 /6 1/2 45 /4 1/ 1/ 60 /3 90 /2 1 0 180 0 1 270 3/2 1 0 360 2 0 1 q 3/ 2 2 2 3/ 2
1/2 The common angles < 90 listed above will become reference angles. Problem 3. Using the table, find: tan p p and cot 6 4 Using the definition of the sine and cosine functions on the unit circle we can find the signs of the trigonometric functions in quadrant. each cosq = x and sin q = y S A T C The above graph shows the results. Problems 4 through 8: 4. In which of the four quadrants is the sine function positive? 5. In which of the four quadrants is the secant function negative? 6. In which of the four quadrants is the cosecant function positive and the cosine function negative? 7. In which of the four quadrants do the tangent function and the cotangent function have the same signs? 7p ^ 7p ^ 7p ^ 8. Find the signs of sin ~, cos ~ and tan ~ . 6 6 6 Recall: cosq = x and sin q = y We have used the definitions to find the sine and cosine of common reference angles. We have used the definitions to find the signs of the trigonometric functions in each quadrant.
We can now use these definitions to evaluate the trigonometric functions of multiples of common reference angles. 2p . q is an angle in Quadrant II. 3 We will define a Reference Triangle. A Reference Triangle is a Right Triangle formed by dropping a perpendicular line from the point, P, to the x axis. (Recall P is the point of intersection of the terminal side of q and the unit circle.) Example 3: Consider q = The blue triangle is The Reference Triangle. We call the acute angle at (0,0) within the triangle, f , the reference angle. f is closely related to q : The sine and cosine of q have the same magnitude as the sine and cosine of f . Only their signs may vary. 2p p In this example, we "see" that f = p  q = p = . 3 3 p ^ p ^ 1 3 and cos ~ = . We know sin ~ = 3 2 3 2 We know the sine function is positive in Quadrant II and the cosine function is negative in Quadrant II. 2p ^ 2p ^ 3 1 and cos ~ = Therefore: sin ~ = + 3 3 2 2 9p . 4 What quadrant is this in? How can I find out? 9p p p 9p = 2p + = one complete revolution and more. \ is in Quadrant I. 4 4 4 4 The Reference Triangle is always a Right Triangle formed by dropping a perpendicular line from the point, P, to the x axis.
Example 4: Consider The blue triangle is The Reference Triangle. We always call the acute angle at (0,0) within the triangle, f , the reference angle. f is always closely related to q : The sine and cosine of q have the same magnitude as the sine and cosine of f . Only their signs vary. may 9p p In the above example, we "see" that f = q  2p =  2p = . 4 4 p ^ 1 p ^ 1 and cos ~ = We know sin ~ = . 4 4 2 2 We know the sine function is positive in Quadrant I and the cosine function is positive in Quadrant I. 9p ^ 9p ^ 1 1 and cos ~ = + Therefore: sin ~ = + . 4 4 2 2 Problem 9. Use the method above to solve the following: 3p ^ a. sin  ~ 4 29p ^ b. cos ~ 6 19p ^ f. cot ~ 3 c. tan(420) 9p ^ d. sec  ~ 4 e. csc(510) We can use this method to solve simple trigonometric equations. Example 5: Solve the following for q where 0 q 2p : cosq = We know the solution to: cos f =
3 2 3 p is f = . 2 6 We also know that the cosine function is positive in quadrants I and IV. Therefore our reference angle looks like this: And our solutions look like this: So: q1 = f = p p 12p p 11p and q2 = 2p  f = 2p  =  = 6 6 6 6 6 3 There is another method of solving: cosq = . 2 3 We can graph y = cosq and y = on the 2 same set of axes and find their points of intersection. p We "see" the first point of intersection is: q1 = 6 11p and the second point of intersection is: q2 = 6 We will commonly use the first method as it is more useful for Calculus. Example 6: Solve the following for q where 0 q 2p : sin q = We know the solution to: sinq = + 3 2 3 p is f = . 2 3 We also know that the sine function is negative in quadrants III and IV. Therefore our reference angle looks like this: And our solutions look like this: So: q1 = p + f = 4p 5p and q2 = 2p  f = 3 3 NOTE: Never solve this and similar problems by plugging a negative number into your calculator. With the sine function, you will get a negative angle (IV quadrant on the unit circle). With the cosine function, you will get an angle in quadrant II only. Your calculator is set up to find the inverse trigonometric functions. This is NOT what we want in these problems. To summarize this method of solving simple trigonometric equations: p 1. Locate f a small positive angle between 0 and . 2 2. Place f in the quadrants corresponding to the given equation. 3. Find a q in the appropriate quadrants. 3 Problem 10. Given: cosq = and sinq is negative . 2 Find: the quadrant of q and sec q . Problem 11. Solve for q where 0 q 2p in the following problems. 1 1 a. cos q =  b. tanq =  3 c. sin q = 2 2 Example 7: In Example 6, Solve the following for q where 0 q < 2p : sin q =  3 , we found 2 q1 = 4p 5p . How does our answer change if the question asks us to solve for all q ? and q 2 = 3 3 We know that the function is periodic. It repeats every 2p radians. Then our solutions should also repeat sine every 2p radians. Our solutions become: 4p 4p 4p 4p q1 = , 2p , (2)2p , (3)2p . . . 3 3 3 3 \ in general q = 4 p + (n)2p where n is an integer 1 3 and 5p 5p 5p 5p , 2p , (2)2p , (3)2p . . . 3 3 3 3 5p \ in general q 2 = + (n)2p where n is an integer 3 q2 = Problem 12. Find all the values for q from Example 5, cosq = 3 2 Example 8: Solve cos2p = 0 for x in (1,1) . x 3p p p 5p np We know cos f = 0 when f = . \ q = , , . . . where n is an odd integer . 2 2 2 2 2 np n 1 3 So p x = 2 fi x = where n is an odd integer. And for x in (1,1) the solution is x = , . 2 4 4 4 Problem 13. Solve the following for x on the given intervals. a. sin2p x = 0 for x in (0,2) c. cos b. tan p 1 x =for x in (1,4) 2 3 p x = 1 for x in (7,7) 3 Identities: sin2 x + cos 2 x = 1 divide by cos 2 x : tan2 x + divide by sin 2 x : 1 1 = sec 2 x + cot 2 x = csc 2 x Double Angle Formulas : sin2x = 2sin x cos x Half  Angle Formulas : sin2 x = 1 cos2x 2 cos 2 x = 1+ cos2x 2 cos2x = cos 2 x  sin2 x = 2cos 2 x 1 = 1 2sin 2 x ...
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This note was uploaded on 10/16/2009 for the course MATH 1224 taught by Professor Dontremember during the Spring '08 term at Virginia Tech.
 Spring '08
 DONTREMEMBER
 Geometry, Unit Circle

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