MATH 155 Midterm: MATH 155 NIU Solution

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turquoisegnu128

15 Feb 2019

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For the next two problems: sin a = 1. [5] find the exact value of cos(a b): [6] find the exact value of sin(a b): [9] find all angles in the range 0 360 for which sin( ) = 3 cos( ). [11] solve sin2( ) + 3 sin( ) = 2. [12] i want to construct a triangle with sides of length a, b, c opposite angles a, b, respectively. If i want a = 4, b = 5 and c = 12: either nd cos(a) or state that no such triangle exists . [13] i want to construct a triangle with sides of length a, b, c opposite angles a, b, respectively. If i want a = 4, b = 5 and c = 2:either nd cos(a) or state that no such triangle exists . [14] i want to construct a triangle with sides of length a, b, c opposite angles a, b, respectively.

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[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

Donât understand why the r(t) in part(a) is <sint, cost> and in part(b) is <cost, sint>. Thank you.

MATH 155 Midterm: MATH 155 NIU Solution

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