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Again we must use sin 1 sin θ θ in addition we must

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Again, we must use sin -1 (sin( θ )) = θ . In addition, we must put some of our wonderful identities to work. Here, we need both sin(2 θ ) = 2sin( θ )cos( θ ) and the beloved pythagorean varmint sin 2 ( θ ) + cos 2 ( θ ) = 1. sin( 2sin -1 (v)) = 2sin(sin -1 (v))cos(sin -1 (v)) //Let θ = sin -1 (v) in in the double angle thingy.// = 2v(1-v 2 ) 1/2 //Set θ = sin -1 (v) in the pythagorean beast and solve for cos(sin -1 (v)), all the while keeping in mind that cosine is non- negative in the interval [- π /2, π /2], since we have - π /2 sin -1 (v) π /2 provided -1 v 1.// 7. (5 pts.) Find the exact value of tan(2 tan -1 (3/4)). tan(2 tan -1 (3/4)) = 2tan(tan -1 (3/4))/[1 - tan 2 (tan -1 (3/4))] = (6/4)/[1 - (9/16)] = 24/7. ...same theme, another variation... 8. (5 pts.)(a) Obtain all solutions to the equation below, and then (b) list the solutions θ with 0 θ < 2 π . sin(3 θ ) = -1 (a) The given equation is equivalent to 3 θ = 3 π /2 + 2k π , for k any integer. Solving this for θ yields θ = π /2 + (2 π /3)k for k any integer. (b) The solutions in the desired interval are π /2, 7 π /6, 11 π /6.
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TEST-03/MAC1114 Page 3 of 4 9. (10 pts.) A right triangle has one angle of 35° and one leg of length 100 meters. What are the two possible lengths for the hypotenuse?? [You may want to sketch the two situations.] Depending on whether the leg is adjacent or opposite the angle, we have either cos(35°) = 100/h or sin(35°) = 100/h. Thus, either h = 100/cos(35°) 122.08 meters or h = 100/sin(35°) 174.34 meters. 10. (5 pts.) A triangle has two sides with lengths 5 feet and 8 feet. If the two sides meet in an angle of 30°, what is the exact length of the third side?? If x denotes the length of the third side, then from the Law of Cosines, we have x 2 = 8 2 + 5 2 - 2(8)(5)(cos(30°) = 89 - 40 3 1/2 . Thus, x = (89 - 40 3 1/2 ) 1/2 , exactly. Note that x 4.44 feet.
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