Exam 3: December 3 rd 7:00-8:30

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MTH 111 - Fall 01 Exam Review (Solutions) Exam : December rd 7:00-8:0 Name: This exam review contains questions similar to those you should expect to see on Exam. The questions included in this review, however, are by no means exhaustive. You should continue to read the textbook, attend SI sessions, consult your instructor, and work on the suggested problems listed on the syllabus for more exam preparation. Good luck! 1. Suppose that an angle α has measure π and an angle β has measure 8π. Are α and β coterminal? We must check to see if α and β differ in measure by an integer multiple of π. Solve the following equation for k: m(α) + πk m(β) π + πk 8π πk 8π π 9π π k But then since k / Z (k is not an integer) we have that α and β are not coterminal.. In which quadrant does an angle with measure 00 lie? Since 00 0 0, angles with measures 00 and 0 are coterminal. Since 0 is between 180 and 70, the angle is in Quadrant III.. Convert 00 to radians. ( ) 00 00 degrees π radians 180 degrees 00π 180 rad 10π rad 5π rad.. Convert 7π 9 to degrees. 7π 9 rad 7 π 9 rad ( ) 180 deg π rad 7 180 deg 7 0 deg 10. 9 5. Find the exact value of the expression cos ( π ) sin ( π ). (Remember that cos (x) means the same thing as (cos(x)).) Using the unit circle, we simply calculate the following: cos ( π ) ( ) ( π sin cos ( )) π ( sin ( )) π ( ) ( ) 1 1 1.

. Find the exact value of the expression cos(π) + sin( π ). (It may be useful to note that π is coterminal with π : this angle is in Quadrant III). cos(π) + sin( π ) cos(π) + sin( π ) 1. 7. Given that sin α 8 17 and α is an angle in Quadrant III, find cos α. Recall the Pythagorean (Fundamental) Identity: cos (x) + sin (x) 1. cos (α) + sin (α) 1 cos (α) + ( 8 17 ) 1 cos (α) + 89 1 cos (α) 89 89 89 cos (α) 5 89 cos(α) ±15 17. However, in Quadrant III, cos x < 0, so we must select the negative value: cos(α) 15 17. 8. Find the amplitude, period, phase shift, and vertical shift (displacement) of the sinusoid y cos(x + π ) 1. We must write our sinusoid in the form y A sin(b(x C)) + D, so we need to factor out of the interior of the function: y cos ( x + π ) 1 y cos ( ( x + π )) ( ( ( π ))) 1 y cos x 1. A general sinusoid has amplitude A, phase shift π B, phase shift C, and vertical shift D. We thus identify the following values: Amplitude Period π B π π Phase shift π Vertical shift 1 9. Evaluate sec(π). sec(π) 1 cos(π) 1 1 1. 10. Evaluate csc( π ). csc ( π ) 1 sin( π ) 1 ( ).

11. Evaluate cot( π ). cot ( π ) cos( π ) sin( π ) 1 1 1 1 ( ). 1. Evaluate tan( π ). tan ( π ) sin( π ) 1 cos( π ) 1 1. 1. What is the period of the function f(x) tan(x) + 1? A general tangent function y tan(bx) has period π B. Thus the function f(x) tan(x) + 1 has period π. 1. What is the period of the function g(x) csc(x)? A general cosecant function y csc(bx) has period π B. Thus the function g(x) csc(x) has period π. 15. Which of the following is not a vertical asymptote of the function f(x) tan(x)? A. x π B. x 0 C. x π D. x 5π Since f(x) tan(x) sin(x), this function has vertical asymptotes wherever cos(x) 0. The cos(x) set of all such x is given by {x R : x π + πk, k Z}. We can generate a few elements in this list with some varying values of k: {..., 5π, π, π, π, π, 5π,...}. Clearly 0 does not appear in this list (while options A, C, and D do), so x 0 is not a vertical asymptote of f(x) tan(x). 1. Evaluate arctan( 1). Setting arctan( 1) x is equivalent to finding the value of x ( π, π ) such that tan(x) 1. Observe that tan(x) 1 sin(x) 1 sin(x) cos(x). cos(x) The only point at which sin(x) cos(x) in Quadrant I or IV is (, ), which is the point of intersection of the unit circle with the terminal side of an angle with measure π.

Thus arctan( 1) π. 17. Find the exact value of the composition sin 1 ( sin( π )). (Be careful!) Recall that sin 1 (sin(x)) x only when x [ π, π ]. Since π is not in this interval, we cannot immediately conclude that this angle is our answer indeed it is not! Evaluate this composition as you normally would. sin( π ) Find the angle x such that sin(x), so ( sin 1 sin( π )) sin 1 ( ). ( )) π π.. Since x π satisfies the equation, sin 1 (sin 18. Find the exact value of the composition tan ( csc 1 ( ) ). First find csc 1 ( ). By an identity, csc 1 ( ) sin 1 ( 1 ). Now sin 1 ( 1 ) π. But then we have that tan ( csc 1 ( ) ) tan ( π ) sin( π ) cos( π ) 1 1 1. 19. Find the exact value of the composition tan 1 ( cot ( )) π. First, realize that cot ( π ) cos ( ) π sin ( π ) 1. Now find the angle x ( π, π ) such that tan(x) 1. Since x π ( )) π that tan (cot 1 π. fills the bill, we have 0. Find the exact value of the sine of the angle α in standard position whose terminal side passes through (, 5). For an arbitrary point (x, y) (off the origin) on the terminal side of an angle α (in standard position), we know that sin(α) y r, where r x + y. Since x and y 5, r ( ) + (5) 1 + 5 1. But then sin(α) y r 5 1 5 1 1. 1. Draw a right triangle having legs of length and. Let β be the angle opposite the shortest side (of length ). Find sec β. By the Pythagorean Theorem, we can solve for the length of the hypotenuse, which we will call

c. Now () + () c 0 c c 0 5. Using the definition of cos, we obtain the following: cos(β) adjacent hypotenuse 5, 5 but since sec(β) 1 cos(β), we simply find the reciprocal of the fraction above. Thus sec(β) 5.. A person stands 0 feet from a maple tree and sights the top with an angle of elevation of 5. How tall is the tree? (Express your answer in exact form.) Draw a right triangle in which the acute angle α of measure 5 is opposite the leg of unknown length (the tree height) and is adjacent to the side of length 0 ft. Label the unknown height as h. Since tan α opposite adjacent, we have that tan(5 ) h 0 h 0 tan(5 ) ft. (For those curious, the tree is roughly 85.780 ft. tall.) ( ) sin( x). Use identities to fully simplify + 1. cos( x) By the even/odd identities, we have that sin( x) ( sin(x)) sin(x) and cos( x) cos(x). Combining this with one of the Pythagorean Identities, we have that ( ) sin( x) + 1 cos( x) ( ) sin(x) + 1 cos(x) (tan(x)) + 1 tan (x) + 1 sec (x).. Find the exact value of sin(105 ). Realizing that 105 0 + 5, we apply the sum identity for the sine function, sin(α + β) sin α cos β + cos α sin β: sin(105 ) sin(0 + 5 ) sin(0 ) cos(5 ) + cos(0 ) sin(5 ) + 1 +.

5. Use identities to fully simplify sin x sec x + cos x csc x. Use the definitions of sec x and csc x to rewrite the above expression as the sum of fractions. Then, get a common denominator and simplify (using the Fundamental Pythagorean Identity at the end). 1 sin x sec x + cos x csc x sin x cos x + cos x 1 sin x sin x ( ) sin x cos x + cos x ( cos x ) sin x sin x cos x (Note that an equivalent expression is csc(x). ) Good luck on Exam #! sin x + cos x sin x cos x 1 sin x cos x 1 sin x 1 csc x sec x. cos x