Review: Know how to find trig functions of 30, 45, and 60.

Transcription

1 Review: Know how to find trig functions of 0, 45, and 60. Always make sure you know the basic triangles below, from which you should easily figure out the information in the table at the right. π θ sin(θ) cos(θ) tan(θ) 6 = 0 π 4 = 45 π = M9500 Precalculus Chapter 7. Trigonometric identities

2 Basic trigonometric identities In the diagram, the terminal side of angle θ goes from the origin to P (x, y). The distance from P to the origin is r. By the Pythagorean theorem, x + y = r. The six trig functions are: cos(θ) = x r sin(θ) = y r tan(θ) = y x sec(θ) = r x csc(θ) = r y cot(θ) = x y It follows from the above that all trig functions can be expressed in terms of sin(θ) and cos(θ) as follows. tan(θ) = sin(θ) cos(θ) cot(θ) = tan(θ) = cos(θ) sin(θ) sec(θ) = cos(θ) csc(θ) = sin(θ) cos(θ) = sec(θ) sin(θ) = csc(θ) Alternate forms: 5 y 4 r = 5 P(4,) y = x = Dividing x + y = r by r gives ( x r ) + ( y r ) = so cos (θ) + sin (θ) = Dividing by cos (θ) gives + sin (θ) cos (θ) = cos (θ) and so + tan (θ) = sec (θ) Dividing by sin (θ) gives cos (θ) sin (θ) + = sin (θ) and so + cot (θ) = csc (θ) M9500 Precalculus Chapter 7. Trigonometric identities θ x

3 Since we derived the last three identities from the Pythagorean Theorem, they are called the Pythagorean trigonometric identities. The word identity means that the expressions on the two sides of the equals sign are equal (identical) for every angle θ. Each of the Pythagorean identities relates two trig functions. You can express each one in terms of the other. You could rewrite cos θ + sin θ = as cos (θ) = sin θ or as sin θ = cos θ = You could rewrite + tan θ = sec θ as tan θ = sec θ or as sec θ tan θ = You could rewrite + cot θ = csc θ as cot θ = csc θ or as csc θ tan θ = You should memorize only the first column of identities, but you need to be prepared to convert any identity in that column to the corresponding identities in the other columns. M9500 Precalculus Chapter 7. Trigonometric identities

4 We sometimes use the following even-odd identities. cos( θ) = cos θ and sin( θ) = sin θ M9500 Precalculus Chapter 7. Trigonometric identities

5 Y 5 We sometimes use the following even-odd identities. cos( θ) = cos θ and sin( θ) = sin θ To see why these are true, look at the diagram. The endline of θ goes from the origin to point (x, y) = (4, ) in Quadrant. cos θ = x r = 4 5 and sin θ = y r = r = 5 P (x, y) = (4, ) θ X M9500 Precalculus Chapter 7. Trigonometric identities

6 Y 5 We sometimes use the following even-odd identities. cos( θ) = cos θ and sin( θ) = sin θ To see why these are true, look at the diagram. The endline of θ goes from the origin to point (x, y) = (4, ) in Quadrant. cos θ = x r = 4 5 and sin θ = y r = 5 The endline of angle θ goes from the origin to point (x, y) = (4, ) in Quadrant 4. Then cos( θ) = x r = 4 5 = cos(θ) and sin( θ) = y r = y r = 5 = sin(θ) It s easy to check (same argument) that tan( θ) = tan θ P (x, y) = (4, ) r = 5 θ θ X r = 5 Q(x, y) = (4, ) M9500 Precalculus Chapter 7. Trigonometric identities

7 Even and odd functions A function y = f(x) is even if for all x in its domain, f( x) = f(x). The graph remains unchanged when it is reflected through the y-axis. Such a graph is also called y-axis symmetric. If point (x, y) is on the graph, then so is point ( x, y). odd if for all x in its domain, f( x) = f(x). The graph remains unchanged when it is reflected through the origin. Such a graph is also called origin symmetric. If point (x, y) is on the graph, then so is point ( x, y). We showed on the last slide that cosine is an even function, while sine and tangent are odd functions. To better understand what this means, look at the graphs below. The function cos(x) is even: for all x, cos( x) = cos(x). Its graph is y-axis symmetric. 0 If point (x, y) is on the graph, so is point ( x, y) - Y X π π π π ( x, y) (x, y) M9500 Precalculus Chapter 7. Trigonometric identities

8 The function sin(x) is odd: for all x, sin( x) = sin(x). Its graph is origin symmetric. If point (x, y) is on the graph, so Y (x, y) 0 X π π π π ( x, y) is point ( x, y) - Y The function tan(x) is odd: for all x, tan( x) = tan(x). Its graph is origin symmetric. If point (x, y) is on the graph, so is point ( x, y) (x, y) 0 X π π π π - ( x, y) - - M9500 Precalculus Chapter 7. Trigonometric identities

9 Simplifying trigonometric expressions Sometimes it is important to transform one trig expression into another. To do so, you need to build new identities. Method : Rewrite expressions in terms of sine and cosine Example : Simplify and rewrite cos θ( + tan θ) in terms of sine and cosine. Solution: Original expression is cos θ( + tan θ) Use + tan θ = sec θ = cos θ sec θ Use cos θ = sec θ = cos θ ( cos θ ) Simplify = Example : Simplify and rewrite sec θ in terms of sine and cosine. sec θ Solution: sec Original expression is θ sec θ Separate into two fractions = sec θ sec θ sec θ Use cos θ = sec θ = cos θ Use sin θ = cos θ = sin θ M9500 Precalculus Chapter 7. Trigonometric identities

10 Sometimes the instruction is more general: Example : Combine fractions and simplify +sin u Solution: +sin u Original expression is Add fractions Expand numerator Use sin u + cos u = = Rewrite numerator = cos u + cos u +sin u cos u + cos u +sin u (+sin u)(+sin u)+cos u cos u = cos u(+sin u) = + sin u+sin u+cos u cos u(+sin u) + sin u+ cos u(+sin u) (+sin u) cos u(+sin u) Cancel = cos u = sec u The last step was a required simplification, since it removed a fraction from the answer. M9500 Precalculus Chapter 7. Trigonometric identities

11 In the next example, we use the important principle that a fraction can be rewritten as its numerator times the reciprocal of its denominator: 4 = 4 and u v = u v Example 4: Combine fractions and simplify cos u sec u + sin u csc u Solution: Since sec u = cos u, then sec u cos u = and so cos u = Since csc u = sin u Original expression is sec u., then csc u sin u = and so sin u = csc u. cos u sec u + sin u csc u Rewrite fractions cos u sec u + sin u csc u Use remarks above = cos u cos u + sin u sin u Pythagorean identity = cos u + sin u = M9500 Precalculus Chapter 7. Trigonometric identities

12 In the next example, we first create a complex fraction. We then simplify the fraction by multiplying its numerator and denominator by the LCD of the nested (little) fractions that it contains. sin u+ csc u+ Example 5: Simplify Solution: sin u+ Original expression is csc u+ sin u+ Rewrite csc u = Multiply top and bottom by sin u Multiply out on bottom Cancel: (since + sin u = sin u + ) = = sin u + sin u(sin u+) sin u( sin u +) sin u(sin u+) +sin u = sin u M9500 Precalculus Chapter 7. Trigonometric identities

13 Example 6 : Simplify +cot u csc u Solution: (the hard way) Original expression is Rewrite cot u and csc u Square the fractions Multiply top and bottom by sin u Use cos u = sin u = = +cot u csc u cos u +( sin u ) u ( ) sin u cos + u sin u sin u = sin u+cos u = sin u+ sin u Cancel: (since + sin u = sin u + ) = sin u + M9500 Precalculus Chapter 7. Trigonometric identities

14 Example 7 : Simplify +cot u csc u Solution: Hint: think about the identity csc u = cot u + Original expression is Use the hint, but rewrite the identity as csc u = cot u Rewrite the numerator Rewrite the numerator +cot u csc u = +csc u csc u = +csc u csc u Split the fraction in two = csc u + csc u csc u Use csc u = sin u = sin u + M9500 Precalculus Chapter 7. Trigonometric identities

15 Quiz Review Example : Simplify and rewrite cos θ( + tan θ) in terms of sine and cosine. Example : Simplify and rewrite sec θ sec θ in terms of sine and cosine. Example : Combine fractions and simplify +sin u cos u + cos u +sin u Example 4: Combine fractions and simplify cos u sec u + sin u csc u Example 5: Simplify sin u+ csc u+ Example 6 : Simplify +cot u csc u Example 7 : Simplify +cot u csc u (use complex fraction simplification) (use a trig identity involving cot u and csc u.) M9500 Precalculus Chapter 7. Trigonometric identities