MIDTERM 3 SOLUTIONS (CHAPTER 4) INTRODUCTION TO TRIGONOMETRY; MATH 141 SPRING 2018 KUNIYUKI 150 POINTS TOTAL: 30 FOR PART 1, AND 120 FOR PART 2

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1 MIDTERM SOLUTIONS (CHAPTER 4) INTRODUCTION TO TRIGONOMETRY; MATH 4 SPRING 08 KUNIYUKI 50 POINTS TOTAL: 0 FOR PART, AND 0 FOR PART PART : USING SCIENTIFIC CALCULATORS (0 PTS.) ( ) = 0., where 0 θ < 0. Give your answers in degrees and in solution set form, and round them off to the nearest tenth of a degree (that is, to one decimal place). (5 points) ) Give the solutions for sin θ In degrees, sin 0. specified interval, 0, 0 ( ).9. Beware that this degree measure is not in the ), so this is not one of our solutions. One solution will be its coterminal twin angle, about.9 + 0, or.. This is in Quadrant IV. The only other Quadrant in which sin( θ ) < 0 is Quadrant III. Another solution is the brother or coreference angle in Quadrant III that is in 0, 0 ). Remember that brothers or coreference angles have the same reference angle; in this case, it is about.9. The desired brother in Quadrant III is given by: about , or.9. (The intersection points have the same y-coordinate, 0..) The approximate solution set is {.9,. }. ) Give the solutions for tan( θ ) =, where 0 θ <. Give your answers in radians and in solution set form, and round them off to the nearest thousandth of a radian (that is, to three decimal places). (5 points) In radians (the assumed measure), tan ( ).07, giving us a Quadrant IV angle. This is not in 0, ), so we add to get a coterminal twin angle in 0, ). One of our solutions is about: [radians]. It is also in Quadrant IV. The only other Quadrant in which tan θ ( ) < 0 is Quadrant II. Another solution is the brother or coreference angle in Quadrant II that is in 0, ). It is about: [radians]. (The indicated terminal sides have the same slope,.) The approximate solution set is {.04, 5.7}.

2 ) A 5-foot ladder is leaning against a tall vertical building that is standing upright and perpendicular from the flat ground. The angle of elevation of the ladder from the ground is 8. How far is the bottom of the ladder from the bottom of the building? Round off your answer in decimal form to three significant digits. (8 points) Let x = the distance from the bottom of the ladder to the bottom of the building. By SOH-CAH-TOA, ( ) = x ( ) = x cos 8! 5cos 8! 5 x.8 feet This seems reasonable, based on the figure. The bottom of the ladder is about.8 feet from the bottom of the building. x 4) Write tan sin, or tan arcsin x, as an equivalent algebraic expression, as in class. Assume x is in the domain of the expression. (7 points) x Let θ = sin, or θ = arcsin x. Then, sin( θ ) = x and θ is in,. We may assume θ is acute when using a right triangle model and SOH-CAH-TOA. We use the Pythagorean Theorem to find the missing (blue) side length, b. ( x) + ( b) = ( ) By SOH-CAH-TOA, tan( θ ) = x + b = 9 b = 9 x x 9 x. ( ) b = 9 x Take the nonnegative root. Warning: 9 x is not equivalent to x. (We usually don t rationalize the denominator when the radicand is variable.) ( ) ; use the graph of y = sin( x) as a guide. (5 pts.) ( ) is graphed in red and blue, with vertical asymptotes (VAs) dashed in brown. 5) Graph two cycles of y = csc x y = csc x

3 PART : NO CALCULATORS ALLOWED! (0 POINTS) ) A central angle of a circle intercepts an arc of length inches along the circle. If the angle s measure is 7 radians, what is the radius of the circle? (4 points) Arc length s = rθ = ( r) ( 7) r = 7 in The radius of the circle is 7 in. 7) Fill out the table below. Rationalize denominators and simplify wherever appropriate. You do not have to show work. (8 points total) θ 0 sin θ ( ) cos( θ ) tan( θ ) sec( θ ) 0 = 0 0 = 0 = = / / = = 4 4 = 0 xxx xxxxxxxxxxx xxxxxxxxxxx xxxxxxxxxxx xxxxxxxxxxx 0 For the special angles from 0 to : Cofunction IDs: Reverse/flip the sin θ Quotient ID: tan θ Use the unit circle for ( ) = sin ( θ ) cos( θ ) (on the left) and = or / = / / = / = 0 : und. 0 : und. 0 : und. 0 : und. ( ) column to get the cos( θ ) column. Reciprocal ID: sec( θ ) = cos( θ ) (on the right, plus comments): Reference angle: Quadrant IV By ASTC, cosine and secant are positive in value, but the others are negative.

4 8) Convert 7 8 radians into degrees. ( points) ( 0! ) " ( ) " 7 8 = ! = 7 80! # 8 = 70 o # ( ) ( ) 9) Fill out the table below. Use interval form (the one with parentheses and/or brackets) except where indicated. You do not have to show work. (4 points) ( ) Domain Range ( ) (, ), f x sin x tan( x) csc( x) sec( x) Use set-builder form. x x + n ( n ) Use set-builder form. { x x n ( n ) } Use set-builder form. x x + n ( n ) cos ( x), tan ( x), (, ) (,, ) (,, 0, ( ), 0) Evaluate the following. (A table you filled out earlier may help.) Rationalize denominators wherever appropriate. ( points total) a) sec 4 (5 points) is outside the interval 0, 4 ), so let s find a more recognizable coterminal twin angle. It will have the same secant value. = = 4 The reference angle for 4 is 4., which is in 0, ). ) cos 4 = or, so sec 4 = cos 4 = / =. See 7).

5 Therefore, sec 4 and sec 4 are either both or both, since 4 and 4 are brothers (or coreference angles ) of 4. and are special Quadrant II angles; is less than By ASTC (or the Unit Circle), cosine and secant are negative in Quadrant II. Answer:. b) tan( 40 ) ( points) The reference angle for 40 is 0, because: 40 is in the interval 0, 0 ) ; among 0, 80, or 0, 80 is closest to 40 ; and = 0. tan 0 ( ) = tan =. (Think: Steep à High slope. ) ( ) is either or, since 40 is a brother or Therefore, tan 40 coreference angle of is a Quadrant III angle, because 80 < 40 < 70. By ASTC (or the Unit Circle), tangent is positive in Quadrant III. Answer:. Note: You may also work in radians. 40 is 4 reference angle and is, again, in Quadrant III. c) csc 4 ( ) ( points) ( ) = 0, so csc( 0) ( ) is also undefined. 4 is coterminal with 0. sin 0 Therefore, csc 4 radians, which has as a is undefined. ) Write the three Pythagorean Identities, as given in class. ( points) sin ( x) + cos ( x) = tan + cot ( x) = csc ( x) ( x) + = sec ( x)

6 ) Which is greater? Box in one: tan( 40 )! vs. tan( 50 )! ( points) tan θ ( ) increases on the θ -interval,, or ( 90!, 90! ). Also, the blue terminal side corresponding to the 50 standard angle has a greater slope (tangent value) than the red terminal side for the 40 standard angle does. ) If sin( θ ) =, find csc( θ ). ( points) 4 csc( θ ) = csc( θ ) ( cosecant is an odd function) = sin( θ ) = / 4 = 4 4) The equation y = 5cos x is equivalent to (Box in one:). ( points) y = 5cos x + y = 5cos x + 5cos x = 5cos x + = 5cos x + 5) Graph one cycle of y = tan 4x because cosine is even. ( ) using the frame provided. Simplify and clearly label all key x- and y-coordinates next to each corresponding grid line. Superimpose the x- and y-axes. If you do not use the frame, make sure you provide all required information. (0 points) Observe: b = 4 > 0 Pivot, P: 0, 0 ( ) ; at frame center for tangent Cycle shape: for tangent, a = < 0 ( falling snake ) Amplitude of Frame (half-height of frame): a = = Period: b = 4 ( ) = 4 Increment: 4 Period Axes: Remember them! 4 =

7 ) Graph one cycle of y = 5sin x using the frame provided. Simplify and clearly label all key x- and y-coordinates next to each corresponding grid line. Superimpose the x- and y-axes. If you do not use the frame, make sure you provide all required information. ( points) Observe: b = > 0 Pivot, P: p, d ( ) = 8, ; see below for p; at frame s left-center Cycle shape: for sine, a = 5 > 0 ( good morning ) Amplitude: a = 5 = 5 Period: b = = Find p, the phase shift: Factoring method: y = 5sin x y = 5sin x y = 5sin x 8 + = p Increment: 4 Period Axes: Remember them! or ( ) = 4 = 4 or 8 Solve x + 4 = 0 : x = 4 x = 8 Period =, so I will accept p = 8 + n ( n ). The phase shift p =. (There is no alternative that is closer to 0.) 8 7) Evaluate the following. ( points total; points each) a) arcsin, which can be written as sin arcsin =, because sin = and,, the range of the arcsine function.

8 ( ( )) ( ( )) =, because, the range of the outside tangent function; b) tan arctan tan arctan is a slope (tangent value). (Thus, is in the domain of the arctangent function.) c) arccos cos 7 arccos cos 7 = 5 because: 5 is the brother or coreference angle of 7 that has the same cosine value as 7 namely, and that is in 0,, the range of the arccosine function. Note: 7 is not in that range, so 7 is not the answer. Also: arccos cos 7 = arccos = 5. 8) Graph y = sin x ( ). Draw in the x- and y-axes, and clearly indicate any endpoints and their corresponding coordinates. ( points)