Alpha Trigonometry Solutions MA National Convention. Answers:

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1 Answers: 1 A C C D 5 A 6 C 7 B 8 A 9 A 10 A 11 C 1 D 1 E 1 B 15 C 16 C 17 D 18 C 19 B 0 C 1 E A C C 5 E 6 B 7 E 8 D 9 D 0 B 1

2 Solutions: 1 A Need to check each answer to 1 k60 and 1 (60 ) = 06. C An even function is one such that f( x) = f(x). As long as one of the functions in the composition is even (namely cosine), the composition is even so II, III and IV are even. C sin(105 ) + sin(15 ) = sin( ) + sin(60 5 ) = sin(60 ) sin(5 ) = ( ) ( ) = 6 D r = sin θ cos θ = sin(θ). Since is even, it will have twice as many or petals. 5 A Call the two quantities A, B. tan(a + B) = tan(a) + tan (B) 1 tan(a) tan (B) = So, A+B is an angle whose tangent is -1. A + B = π. + ( 5) 1 ( ) ( 5) = 1 1 = 1. 6 C By symmetry, the line from P to O, and the lines to the tangent point form a triangle. The short side is the radius =, the hypotenuse is then. 7 B There are π radians in 180, so 8 A 8π 15 (180 π ) = 96 i 1 = cis ( π ) 1 i = cis ( π ) (i 1) 5 (1 i) = 8 cis (5 ( π ) (π )) = 16 cis(π) = 16 9 A The cosine of the angle between the vectors is: <,> <,> = 1 <,> <,> right triangle with an angle of cosine 1/1 is 5. The tangent is The rd side of a 10 A arcsin has a domain of [ 1, 1]. The range of secant is (, 1] [1, ). We need sec(x) = ±1, so cos(x) = ±1; x {kπ k is an integer} 11 C Obviously x=0 is a solution and since both sides are odd functions, we only need to consider one side. The first crossing will be at little less than π and the second and third crossing a little less and a little more than 5π. On the next period, x/10 will be larger than 1. In total, there are positive, 1 zero and negative solutions = 7.

3 1 D The building, shadow, and the light beam forms a right triangle. So, height = tan 50. shadow height = 70 tan 50 1 E Just a change of units. 00 rev π rad min 1 rev 1 min = 80π rad/sec 60 sec 1 B Since the triangle is isosceles, the obtuse angle must be between the two sides of 10. Since the cosine of an obtuse angle is negative, we must have c > a + b = 00. c can be 15, 16, 17, 18 or 19 which sum to C cos ( π 1 ) + sin ( ( π 1 )) = (1 + cos (π 6 ) ) + sin ( π (1 + 6 ) = ) + 1 = = + 16 C A vector orthogonal (perpendicular, normal) to the given plane is <,,1 >. Just have to take dot product with the other normal vectors seeing which gives 0. <,,1 > < 1,,5 >= D sin ( x ) has period 6π and cos (x ) has period π. The LCM(6π, π) = 1π. 18 C Geometric definition of parabola. Let the point be (a,b) and as an example line y=k. Equating squared distances. (x a) + (y b) = (y k) A parabola. 19 B Using a rotation matrix: (x a) = (b k)y + k b cos θ sin θ [ sin θ cos θ ] [ 1/ / ] = [ / 1/ ] [ ] = [ 1 ] Now switch the sign of y. (, 1) 0 C cos θ < 1 and 5 sin θ > 1 means cos θ < 0, sin θ > 0. The angles must terminate in the second quadrant. Only π 1 E The angle between the points measures 5, the Law of Cosines can find the squared distance. d = + 5 ()(5) cos(5 ) = 15

4 A Divide the region into pieces. The sector has area: πr ( A π ) = A r. The triangle has base r, and height r sin A, with area 1 sin A r. The total area is A + sin A ( ) r C r( sin θ + 5 cos θ) = 10, use y = r sin θ and x = r cos θ and simplify. y + 5x = 10 A line so that the length of the graph is just the distance between (0,5) and (,0) or + 5 = 9. C 1 = (sin x + cos x) = sin + sin x cos + cos x = sin x + cos x + sin x = sin x + cos x yielding an answer of E cos ( π 5 ) + cos (π 5 ) + cos (π 5 ) + cos (π ) + cos(π) = cos(π) = 1. 5 We have cos ( π 5 ) = cos (π 5 ), and cos (π 5 ) = cos (π 5 ). 6 B Let: 8x = ()(sin π 16 )(cos π 16 )(cos π 8 )(cos π ) = () sin ( π 8 ) cos (π 8 ) cos (π ) = sin ( π ) cos (π ) = sin ( π ) = 1. x = E The area of the triangle is: A = 1 (ab) sin(c). To maximize, let sin(c) = 1; giving an area of 1 (0)(5) = D The amplitude is (00-100)/=150, the coefficient of sin. The vertical shift will be (00+100)/=50. The coefficient of x is π period = π 6. Finally, since the peak is months into the cycle rather than 1/= months, the phase shift is 1 month. s(x) = 150 sin ( π (x 1))

5 9 D So, sin(θ) = 0 only when θ = kπ. The solutions for sin(x), sin(x) and sin(x) are subsets of the others. sin(x) has solutions: π, π, π, π. sin(5x) adds an additional solutions (1,,, over 5 π), sin(6x) provides more: (1,,, 5 over 6 π) for a total of 1. 0 B Assign the variables: h the height of the tree, x distance from ground Mary to tree and y distance from ground Nancy to tree. Now equations. sin B = h x ; sin C = h y d = x + y xy cos A Solve the first eqns for x and y and substitute into the third. d = ( h sin B ) + ( h sin C ) ( h sin B ) ( h ) cos A sin C Finally, solve for h. d sin B sin C h = sin B + sin C sin B sin C cos A 5

MTH 122: Section 204. Plane Trigonometry. Test 1

MTH 122: Section 204. Plane Trigonometry. Test 1 Section A: No use of calculator is allowed. Show your work and clearly identify your answer. 1. a). Complete the following table. α 0 π/6 π/4 π/3 π/2 π