Math 143 Final Exam Practice Problems - Part 2

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1 1. Find the exact value of the following: (a) sin( π 1 ) sin( 1 ) sin(30 ) Math Final Exam Practice Problems - Part sin(30 ) cos( ) sin( ) cos(30 ) ( 1 ) ( ) ( ) ( 3 ) (b) cos(10 ) cos(60 + ) cos(60 ) cos( ) sin(60 ) sin( ) ( ) ( ) ( 3 ) ( ) Given that csc 1 and cos φ, where is in the third quadrant, and φ is in the fourth quadrant, find exact values of: 3 1 φ (a) cos 3 (b) sin(φ) sin φ cos φ ( )( 1 ) (c) cos(φ ) cosφcos + sin φ sin ( 1 (d) sin( + φ) sin cos φ + sinφcos ( )( 1 )( ) ( ) ( ) ) ( )( ) Verify the following identities by transforming the left hand side into the right hand side: ( ) 3π (a) cos sin ( ) 3π cos cos 3π cos + sin 3π sin 0 cos + ( 1) sin sin 1 (b) 1 sin x sinx tanxsec x 1 1 sin x sinx (1 sin x) 1 + sin x 1 + sinx 1 + sinx 1 sin x cos x sin x tanxsec x. cos x (c) sec t csc tant cott t sin t cos t sec t csc t 1 cos 1 t sin t sin t cos t cos t sin t sin t cos t sin t cos t tant cot t (d) sin 3t cos 3t 1 sin 6t sin 3t cos 3t 1( sin 3t cos 3t) 1 sin 6t. 1 + cos t (e) cott sin t 1 + cos t 1 + cos t sin t 1 sin t + cos t sin t sin t cos t sin t cos t sin t cos t () 1 + cos t + cos t sin t cos t sin t cos t cos t sin t cos t cos t sin t cott.

2 . Find exact solutions to the following equations with 0 < π. (a) sin(x + π ) 3 Then sin(x + π ) 3, or sin 3, where x + π Therefore, x + π π 3 + πk or x + π π 3 + πk But then, x π 1 Thus x π 8 + π Hence x π, 37π, 61π, 8π π + πk or x + πk 1 17π k or x + πk 8 or x 17π 8, 1π 8, 6π 8, 89π 8. (b) cos 3 3 cos Then cos 3 3 cos 0, or cos( cos 3) 0 Thus either cos 0, or cos 3 0, so cos 3, or cos 3, so cos ± 3 Hence π, 3π, or π, π, 7π 11π, and (c) sin t sin t 0 Then sin t cos t sin t 0, so sint( cos t 1) 0. Thus either sint 0, or cost 1. Hence t 0,π, π, and π. 3 3 (d) cos cos 0 Notice this equation is quadratic in form. Therefore, we let u cost. We then have u u 0. Solving this using the quadratic formula: u ± ()( ) () ± 6. Therefore, cos ± 6. So either cos 3.66, or cos.766 Then cos 1 (.766).3 radians, or π radians.. Given the tables below, find the following: x f(x) (a) f 1 () (b) f(g 1 (9)) f(6) (c) g(f 1 ()) g(6) 9 x g(x) Determine whether or not the following functions are one-to-one. You must justify your answer to each part. (a) f(x) 3x f(x) is not one-to-one. Notice that f(1) f( 1) 1.. (b) g(x) x Suppose that g(x 1 ) g(x ) for two x-values x 1 and x. Then x 1 x, but then x x 1, or x x 1. Hence g(x) is a one-to-one function.

3 7. Find the exact value of the following: (a) sin 1 ( 3 ) π 3 (b) tan 1 ( 3) π 3 (c) cos 1 ( π) is undefined. (d) cos(cos 1 ( 1 )) cos(π 3 ) 1 (e) sin 1 (sin( π 3 ) sin 1 ( 3 ) π 3 (f) tan(cos 1 ( 1 )) tan(π 3 ) 3 (g) cos( tan 1 ( 7 )) cos() cos sin, where is described below: 7 7 ( ) ( ) Then cos sin Express tan(cos 1 ( x )) algebraically. x 1 x 1? x 7 7 First, we solve for the missing side in the triangle above using the Pythagorean Theorem in order to obtain: (x 1) x x x 1 x x 3x 1. Thus tan(cos 1 ( x x 1 )) x. x 3x 1

4 9. Solve the following triangles: (these are not neccessarily drawn to scale) (b) (a) 80 o o 0 o (c) 7 0 o (d) For triangle (a), γ , b 17 sin 80 sin 3 9., and c 17 sin 6 sin sin 0 For triangle (b), sinγ.81. Therfore, either γ 3.6, or γ If γ 3.6, then β sin 1., and b.1. sin 0 If γ 1., then β sin 1.6, and b sin 0 For triangle (c), we have that a (1)(7) cos , So a Then cosβ , and β 11.. Thus γ (7)(10.6) For triangle (d), we have that cosα 1 1 (1)().7636, and so α 0.. Similarly, β 1 1 (1)(1).319, and so α Thus γ Find the area of the following triangles o 18 0 For the rightmost triangle, if we drop a perpendiclar to the base, we have h sin 0, or h sin 0. Then since A 1 bh 1 (18)() sin square units. For the other triamgle, we use Heron s formula. s 1 ( ) 31, so A (31)(31 7)(31 0)(31 1)

5 11. Suppose that a regular pentagon inscribed in a circle has sides of length 10cm. Find the area of the pentagon (See the figure below) cm First notice that we can find the area of the pentagon by subdividing it into congruent triangles and then finding the area of one of these triangles. Recall that the central angle of the triangle is 360 7, and the triangle is isosceles, so the base angles are each. Therefore, the base of 10sin the triangle has length 10cm, and the other two sides have length x 8.cm. Then sin 7 the height of the triangle is given by h 8. sin. Hence the area of the triangle is given by A 1(10)(8.) sin 3.38cm. Thus, the area of the pentagon is approximately (3.38) 171.9cm. 1. A rectangular box measures 8 feet by feet by feet. Find the angle between the diagonal on the front of the box with the diagonal on one of the sides of the box. (See the figure below). 8 ft ft ft We use the Pythagorean Theorem in order to find the lengths of the diagonals on the front, top, and side of the box: d f , d s , and d t Next, applying the Law of Cosines to the triangle formed by the three diagonals: cos ( 89)( 1).86, so 6.6. Express the following in the form a + bi. You do not have to use trigonometric forms. (a) (7 i) (6 + 11i) 1 3i (b) (7 i)(6 + 11i) 1i + 77i i 6 + 6i (c) 7 i i 6 11i 6 11i (d) i 36 (i ) i 77i + i 36 11i 0 89i i

6 1. Let z 1 i and z 1 + i (a) Find the trigonometric form of z 1 r +. is best found by plotting z 1 and using common sense. z 1 cis ( ) π (b) Find the trigonometric form of z r tan ( 1 1) + π So z cis ( tan ( ) 1 1) + π. (c) Express (z 1 ) 6 in the form a + bi z1 6 ( 0 ) 6 ( ) cis 6 π 0 3 cis ( ) ( 30π 1, 000cis 3π 1, 000i. (d) Find the fourth roots of z 1 i ) 1, 000(cos 3π + i sin 3π ) s r α 0 6. Thus w 0 8 0cis(6. ), w 1 8 0cis( ) 8 0cis(16. ), w 3 8 0cis( ) 8 0cis(36. ), and w 8 0cis( ) 8 0cis(36. ). Note: Since the material on polar coordinates from Chapter 11 is not required material for the final exam, I am going to omit the assigned problems from this material on the review assignment.