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39 Section 7.5 – Trigonometric Functions: Unit Circle Approach The unit circle is a circle of radius 1 centered at the origin. If we have an angle in standard position superimposed on the unit circle, the terminal side will intersect the unit circle at a particular point. The point of intersection will depend on the value of the angle. This implies that the x-coordinate and the y-coordinate of the point will depend on the value of the angle. (– (– (– 2 2 3 2 , ) ( – (– ,– 2 2 π 1 2 ,– (– 2 2 1 , 2 ) π 2 2π 3 120˚ 135˚ 150˚ 90˚ ( 21 , π 3 60˚ ) – 3 2 ( 45˚ 30˚ 4π 3 330˚ 315˚ 7π 270˚ 300˚ 4 3π 2 ) (0, – 1) 5π 3 ) 2 2 11π 6 ( ( 21 , – 2 2 , ( π 6 0˚ 210˚ 225˚ 5π 240˚ 4 3 2 π 4 180˚ 7π 6 ) (0, 1) 3π 4 5π 6 (– 1, 0) 3 2 3 2 ) 2 2 , 1 2 1 , 2 3 2 , ) 1 2 ) (1, 0) 0 ( 2 2 3 2 3 2 ,– ) ,– 2 2 1 2 ) ) Thus, for an angle that measures 30˚ in standard position, the terminal side will pass through the point ( 3 2 , 1 2 ) on the unit circle. For an angle that measures 270˚ in standard position, the terminal side will pass through the point (0, – 1) on the unit circle. Since the x- and y-coordinates of the point on the unit circle intersected by the terminal side depend on the measure of the angle, we can define each coordinate as a function of the angle. 40 Definition If θ is an angle in standard position and if the terminal side of θ intersects the unit circle at (a, b), then sin(θ) = b (sine of θ) cos(θ) = a (cosine of θ) There are four other trigonometric functions we can define in reference to our unit circle. We can first talk about the slope of the terminal side of θ in standard position. Two points on that line are (0, 0) and (a, b), so the slope b−0 b m= = . This value again depends on the value of θ, so we will a−0 € a define the tangent function (tan(θ)) to be equal to the slope of the terminal side. The other three trigonometric functions, the cosecant (csc(θ)), the secant € (sec(θ)), and the cotangent (cot(θ)),are the reciprocals of the first three trigonometric functions sine, cosine, and tangent. Definition If θ is an angle in standard position and if the terminal side of θ intersects the unit circle at (a, b), then b sin(θ) = b cos(θ) = a tan(θ) = csc(θ) = 1 b sec(θ) = Note that csc(θ) = € 1: Objective 1 , sin(θ) sec(θ) = 1 a 1 , cos(θ) cot(θ) = a a b 1 and cot(θ) = . €tan(θ) € Values of the Trigonometric € Find the Exact Functions Using a Point on the Unit Circle. € € € Let P = (x, y) be the point on the unit circle that corresponds to t. Find the values of the six trigonometric functions of t: ( 23 , – Ex. 1a Solution: a) € 5 3 ) Ex. 1b 5 3 sin(t) = b = – € tan(t) = € € € b a = − 5 3 2 3 € (0, 1) cos(t) = a = =– 5 3 ÷ 2 3 =– 5 3 • € € € € € 2 3 3 2 =– 5 2 41 csc(t) = sec(t) = 1 b = 1 a = 1 − 1 2 3 5 3 = =– 3 5 =– 3 5 5 3 2 € € 2 € xuc 3 2 2 cot(t) = = 3 = • – =– =– y uc 5 3 5 5 − € € 3 € sin(t) = b = 1 cos(t) = a = 0 1 b tan(t) = = which is undefined 0 a € € € € 1 1 € csc(t) = = = 1 € b) ( sec(t) = € cot(t) = € € Objective 2: € b 1 a a b = = 1 1 0 0 1 ) 2 5 5 which is undefined =0 Use a Circle of Radius r to Evaluate the Trigonometric Functions. Now, consider an angle that is in standard position with a terminal side that intersects a circle of radius of r at the point (a, b). By the Pythagorean Theorem, a2 + b2 = r2. Using similar triangles, it can be shown that: b a b sin(θ) = cos(θ) = tan(θ) = csc(θ) = r r b sec(θ) = r r a cot(θ) = a a b €Theorem € € Let θ be an angle in standard position whose terminal side intersects the € € circle x2 + y2 = r2 at the point € (a, b). Then b a b sin(θ) = cos(θ) = tan(θ) = csc(θ) = r r b sec(θ) = r r a cot(θ) = a a b €Find the six trigonometric values € € the following of an angle θ with conditions: € € Ex. 2 (– 5, 12) is a point on€the terminal side. 42 Solution: First, we need to find the radius of the circle: r2 = a 2 + b 2 r2 = (– 5)2 + (12)2 = 25 + 144 = 169 r = 169 = 13 (ignore the – answer) Thus, a = – 5, b = 12, and r = 13: sin(θ) = csc(θ)€= 12 13 13 12 cos(θ) = sec(θ) = −5 13 13 −5 5 13 13 – 5 =– tan(θ) = = cot(θ) = 12 =– −5 −5 =– 12 12 5 5 12 €Objective 3: Domain and € Range € of Trigonometric Functions. € € € € € Let's re-examine the following theorem: € € Theorem Let θ be an angle in standard position whose terminal side intersects the circle x2 + y2 = r2 at the point (a, b). Then b a b sin(θ) = cos(θ) = tan(θ) = csc(θ) = r r b sec(θ) = r r a cot(θ) = a a b of the sine and cosine functions is all real €Since r > 0, then the domain € € numbers. The tangent and secant functions are undefined when a is 0. € € €π π 3π That occurs at θ = , , and every odd multiple of or 90˚. The 2 2 2 cotangent and the cosecant functions are undefined when b = 0. This happens when θ = 0, π, and every multiple of π or 180˚. € €since a ≤ r, and b ≤ r, then both € the sine and the cosine For the range, functions will have a range of between – 1 and 1 inclusively and the secant and cosecant functions will have a range of all values less than or equal to – 1 or greater than or equal to 1. Since a is sometimes less than b, sometimes equal to b , and sometimes larger than b, the range of the tangent and cotangent functions is all real numbers. 43 Function Domain Range sin(θ) (– ∞, ∞) [– 1, 1] cos(θ) (– ∞, ∞) [– 1, 1] tan(θ) sec(θ) csc(θ) cot(θ) {θ| θ ≠ nπ 2 , n is an odd integer} nπ 2 , n is an odd integer} {θ| θ ≠ n(90˚), n is an odd integer} {θ| θ ≠ €{θ| θ ≠ n(90˚), n is an odd integer} {θ| θ ≠ mπ, m is an integer} {θ| θ ≠ n(180˚), n is an integer} € {θ| θ ≠ mπ, m is an integer} {θ| θ ≠ n(180˚), n is an integer} Objective 4: (– ∞, ∞) (– ∞, – 1] U [1, ∞) (– ∞, – 1] U [1, ∞) (– ∞, ∞) Determine the Period of the Trigonometric Function Periodic functions are functions that repeat at regular intervals. If we take the sine and/or cosine of two angles that are coterminal, we will get the same result. This means that the sine and cosine function are periodic functions. This leads to the following definition: Definition A function f is periodic if there is a positive number p such that for any θ in the domain of f, θ + p is also in the domain of f and f(θ + p) = f(θ). The smallest such p value is called the (fundamental) period of f. Since two coterminal angles differ by multiplies of 2π, the smallest such p value would be 2π. Thus, the period of the sine and cosine function is 2π. Since the secant and cosecant functions are reciprocals of the cosine and sine function, then their periods are also 2π. The tangent and cotangent functions are positive when x and y have the same sign, This occurs in the first and third quadrants and thus tan(θ + π) = tan(θ) and cot(θ + π) = cot(θ). Thus, the period of the tangent and cotangent function is π. Periodic Properties: sin(θ + 2π) = sin(θ) csc(θ + 2π) = csc(θ) cos(θ + 2π) = cos(θ) sec(θ + 2π) = sec(θ) tan(θ + π) = tan(θ) cot(θ + π) = cot(θ) 44 Function sin(θ) cos(θ) tan(θ) Period 2π 2π π Function csc(θ) sec(θ) cot(θ) Find the exact value of the following: 25π Ex. 3a sin( ) Ex. 3b cot(7π) Ex. 3c tan( 6 Solution: a) sin( € b) c) 25π 6 ) = sin( Period 2π 2π π π 6 + 2(2π)) = sin( 3 3 π 6 )= 4π 3 ) 1 2 cot(7π) = cot(π + 6(π)) = cot(π) which is undefined. € 7π π π sec( ) = sec( + 2π) = sec( ) = 2 3 € Objective 6: € € € Even-Odd Properties of Trigonometric Functions. € an angle € € with the terminal side passing Let θ be in standard position through the point (a, b). Then the angle – θ will have the terminal side pass through the point (a, – b). (a, b) r θ r (a, – b) Since cos(θ) = =– € b a a r –θ = cos(– θ), then the cosine function is even. But – sin(θ) = sin(– θ), which means that the sine function is odd. Similarly, the secant function is even (reciprocal of the cosine function) and the cosecant function € is odd (reciprocal of the sine function). Now, let's examine the tangent function: tan(– θ) = sin(− θ) cos(− θ) = −sin(θ) cos(θ) = – tan(θ) So, the tangent function is odd and likewise, the cotangent function is also odd. 45 Even-Odd Properties of Trigonometric Functions: sin(– θ) = – sin(θ) cos(– θ) = cos(θ) tan(– θ) = – tan(θ) csc(– θ) = – csc(θ) sec(– θ) = sec(θ) cot(– θ) = – cot(θ) Only the cosine and the secant functions are even; all the other ones are odd. Find the exact value of the following: Ex. 8a cos(– 30˚) Ex. 8b sin(– Ex. 8c csc(– cot(– 240˚) Ex. 8d π ) 2 5π 3 ) Solution: 3 2 a) cos(– 30˚) = cos(30˚) = b) sin(– c) cot(– 240˚) = – cot(240˚) (odd function) 240˚ is in quadrant III, so the reference angle is 240˚ – 180˚ = 60˚. Also, the cotangent is positive in Q III. So, cot(240˚) = cot(60˚) π 2 ) = – sin( π 2 )=–1 (even function) (odd function) Thus, – cot(240˚) = – cot(60˚) = – d) csc(– 5π 3 5π 3 ) = – csc( 5π 3 ) 1 = – tan(60o ) 1 3 3 3 =– (odd function) is in quadrant IV, so the reference angle is 2π – 5π 3 = Since the cosecant is negative in Q IV, then 5π π csc( ) = – csc( ). Hence, 3 – csc( = 2 3 3 5π 3 3 ) = – [– csc( π 3 )] = csc( π 3 )= 1 sin( π 3 ) = 1 3 2 = 2 3 π 3 .