Math 104 Midterm 3 review November 12, 2018

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1 Math 04 Midterm review November, 08 If you want to review in the textbook, here are the relevant sections: 4., 4., 4., 4.4, 4..,.,. 6., 6., 6., , 7., 7., 7.4. Consider a right triangle with base angle α and top angle β. Assume that the side opposite α has length, the side adjacent α has length and hypotenuse has length. Evaluate the following trigonometric functions. (.) sin(α) (.) cos(α) (.) sin(β) (.4) cos(β) (.) tan(α) (.6) tan(β). Suppose 0 α π/ and sin(α) = /. Suppose 0 β π/ and tan(β) = /. Use this information to answer the following questions. (.) If a right triangle has base angle α and hypotenuse 6, what are the lengths of the opposite and adjacent sides? (.) If a right triangle has base angle α and opposite side, what are the lengths of the adjacent side and hypotenuse? (.) If a right triangle has base angle β and hypotenuse 4, what are the lengths of the opposite and adjacent sides? (.4) If a right triangle has base angle β and opposite side 0, what are the lengths of the adjacent side and hypotenuse?. Simplify these trigonometric expressions. (.) cos(tan (/)) (.) tan(cos (/7)) (.) sin(tan (x/)) (.4) tan(sin (/x)) Use these identities to help you solve the next three problems. sin(x) = sin(x) cos(x) cos(x) = cos (x) sin (x) sin (x) = ( cos(x)) cos (x) = ( + cos(x)) sin(α + β) = sin(α) cos(β) + cos(α) sin(β) sin(α β) = sin(α) cos(β) cos(α) sin(β) cos(α + β) = cos(α) cos(β) sin(α) sin(β) cos(α β) = cos(α) cos(β) + sin(α) sin(β) 4. Simplify these expressions. (4.) cos(sin (x/) + tan (/x)) (4.) sin(cos (/x) sin (x)). Assume that α and β are angles in the first quadrant with cos(α) = / and tan(β) = 7. Use the half- and double-angle formulas to evaluate these expressions. (.) cos( α) (.) sin(α) (.) cos( β) (.7) sin(β) (.) sin( α) (.4) cos(α) (.6) sin( β) (.8) cos(β)

2 6. Use the table of values at right to evaluate the following. sine α β (6.) sin(α) (6.) cos(β α) (6.) sin(α + β) (6.4) cos(α) cosine 4 7. Write out the first quadrant of the unit circle. Be sure to include the x- and y-coordinates of the points corresponding to the angles 0, π/6, π/4, π/ and π/. Use your quarter unit circle to fill in these tables of values. (7.) x cos(x) cos(x) 0 π/6 π/4 π/ π/ (7.) x tan(x) tan (x) 0 π/6 π/4 π/ π/ 8. Find all 0 x π/ satisfying each equations below. (8.) 4 sin (x) cos(x) cos(x) = 0 (8.) 4 cos (x) cos(x) = 0 (8.) cos (x) sin(x) sin(x) = 0 (8.4) sin (x) sin(x) = 0 9. Solve these exponential equations for x. (9.) e x + e x 6 = 0 (9.) e x 8e x + = 0 (9.) e x 4e x + e x 4 = 0 (9.4) e x + e x 9e x 8 = 0 0. Simplify these expressions involving logarithms. (0.) ln(/x ) (0.) ln(x /) (0.) log (6) log () (0.4) log (40) log ()

3 Solutions: (.) sin(α) = (.) cos(α) = (.) sin(β) = (.4) cos(β) = (.) tan(α) = (.6) tan(β) = (.) Since sin(α) =, it follows that the basic right triangle with base angle α has opposite side, hypotenuse and adjacent side. Now let y be the length of th eopposite side of the given triangle and x the length of the adjacent side. Since the hypotenuse is 6, it follows that = sin(α) = y 6 and hence y = 4. Similarly, = cos(α) = x 6 and hence x =. (.) Let x be the length of the adjacent side of the given triangle and z the length of the hypotenuse. Since the opposite side has length, it follows that = sin(α) = z and hence z = 9. Likewise, = tan(α) = x and thus x =. (.) Since tan(β) =, it follows that the basic right triangle with base angle β has opposite side of length, adjacent side of length and hypotenuse of length 4. Let x be the length of the adjacent side of the given triangle and y the length of the opposite side. Since the hypotenuse of the given triangle is 4, it follows that 4 = sin(β) = y 4 and hence y =. Similarly, and thus x = 9. 4 = cos(β) = x 4 (.4) Let x be the length of the adjacent side of the given triangle and z the hypotenuse. Since the opposite side has length 0, it follows that = sin(β) = 0 4 z and so z = 4. Likewise, and therefore x = 6. = tan(β) = 0 x

4 (.) cos(tan ( )) = (.) tan(cos ( 7 )) = 4 (.) sin(tan ( x )) = x +x (.4) tan(sin ( x )) = x 9 (4.) cos(sin (x/) + tan (/x)) = cos(sin (x/)) cos(tan (/x)) sin(sin (x/)) sin(tan (/x)) = 4 x x x 9 + x 9 + x (4.) sin(cos (/x) sin (x)) = sin(cos (/x)) cos(sin (x)) cos(cos (/x)) sin(sin (x)) = x 4x x x x (.) cos ( α) = ( + cos( α)) = ( + cos(α)) = ( + ) = Therefore, cos( α) =. (.) sin ( α) = ( cos( α)) = ( cos(α)) = ( ) = Therefore, sin( α) =. (.) First, use the fact that cos(α) = to conclude that sin(α) = 8. Now use the double angle formula. sin(α) = sin(α) cos(α) = 8

5 (.4) cos(α) = cos (α) sin (α) = = 7 9 (.) First of all, use the fact that tan(β) = 7 to conclude that sin(β) = 7 0 and cos(β) = 0. Now use the half angle formula for cosine. It now follows that cos( β) = ( + 0 ) cos( β) = ( + cos( β)) = ( + cos(β)) = ( + 0 ) (.6) sin ( β) = ( cos( β)) Thus it follows that sin( β) = ( 0 ). = ( cos(β)) = ( 0 ) (.7) sin(β) = sin(β) cos(β) = (.8) cos(β) = cos (β) sin (β) = = 48 0 (6.) sin(α) = sin(α + α) = sin(α) cos(α) + cos(α) sin(α) = ( sin(α) cos(α)) cos(α) + (cos (α) sin (α)) sin(α) = ( 4 ) 4 + ( 4 ) = 48 + = 7

6 (6.) cos(β α) = cos(β) cos(α) + sin(β) sin(α) = = (6.) sin(α + β) = sin(α) cos(β) + cos(α) sin(β) = + 4 = + 48 (6.4) I am going to do some preliminary work so there s less mess later. Also, cos(α) = cos (α) sin (α) = 4 = sin(α) = sin(α) cos(α) = 4 = 4 Now compute cos(4α) and sin(4α). and = 4 6 cos(4α) = cos (α) sin (α) = ( ) ( 4 6 ) = = 4 6 sin(4α) = sin(α) cos(α) = 4 6 =

7 It now follows that sin(α) = sin(4α + α) = sin(4α) cos(α) + cos(4α) sin(α) = = 64 (7.) x cos(x) cos(x) 0 - π/6 π/4 π/ 0 π/ 0 (7.) x tan(x) tan (x) π/6 π/4 π/ 6 π/ VA VA (8.) First factor the given equation. 4 sin (x) cos(x) cos(x) = 0 (4 sin (x) ) cos(x) = 0 Thus, there are two cases: either 4 sin (x) = 0 or cos(x) = 0. In the first case, sin (x) = 4 and hence sin(x) = x = π or x = π., i.e., x = π. On the other hand, if cos(x) = 0, then x = π. Thus, either (8.) First factor the given equation. 4 cos (x) cos(x) = 0 (4 cos (x) ) cos(x) = 0 Thus, there are two case: either 4 cos (x) = 0 or cos(x) = 0. In the first case, cos (x) = 4 and hence cos(x) = x = π 6 or x = π., i.e., x = π 6. On the other hand, if cos(x) = 0, then x = π. Thus, either (8.) First factor the given equation. cos (x) sin(x) sin(x) = 0 ( cos (x) ) sin(x) = 0 There are two cases. Either cos (x) = 0 or sin(x) = 0. In the first case, cos (x) = and so cos(x) = or x = 0., i.e., x = π 4. In the second case, sin(x) = 0 and so x = 0. Thus, either x = π 4

8 (8.4) First factor the given equation. sin (x) sin(x) = 0 ( sin (x) ) sin(x) = 0 There are two cases, either sin (x) = 0 or sin(x) = 0. In the first case, sin (x) = and hence sin(x) = x = π 4 or x = 0., i.e., x = π 4. In the second case, sin(x) = 0 and so x = 0. Thus, either (9.) Factor the given equation. e x + e x 6 = 0 (e x + )(e x ) = 0 Thus, either e x = (which is impossible) or e x =, i.e., x = ln(). (9.) Factor the given equation. e x 8e x + = 0 (e x )(e x ) = 0 Thus, either e x = or e x =, i.e., x = ln() or x = ln(). (9.) Factor the given equation by grouping terms. e x 4e x + e x 4 = 0 e x (e x 4) + (e x 4) = 0 (e x + )(e x 4) = 0 Thus, either e x = (which is not possible) or e x = 4, i.e., x = ln(4). (9.4) Factor the given equation by grouping terms. (0.) e x + e x 9e x 8 = 0 e x (e x + ) 9(e x + ) = 0 (e x 9)(e x + ) = 0 (e x )(e x + )(e x + ) = 0 Thus, either e x =, e x =, or e x =. The second and third options are impossible and hence e x =, i.e., x = ln(). ln(/x ) = ln() ln(x ) = ln() ln(x)

9 (0.) ln(x /) = ln(x ) ln() = ln(x) ln() (0.) log (6) log () = log (6/) = log (8) = log ( 4 ) = 4 (0.4) log (40) log () = log (40/) = log (8) = log ( ) =