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., 6.4 7., 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)
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 ()
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
(.) 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
(.4) cos(α) = cos (α) sin (α) = 9 8 9 = 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) = 7 0 0 (.8) cos(β) = cos (β) sin (β) = 0 49 0 = 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.) cos(β α) = cos(β) cos(α) + sin(β) sin(α) = = + 4 4 + (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 ) = 9 6 96 6 = 4 6 sin(4α) = sin(α) cos(α) = 4 6 = 84 6 6
It now follows that sin(α) = sin(4α + α) = sin(4α) cos(α) + cos(4α) sin(α) = 84 6 6 4 + 4 6 = 64 (7.) x cos(x) cos(x) 0 - π/6 π/4 π/ 0 π/ 0 (7.) x tan(x) tan (x) 0 0 0 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
(8.4) 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)
(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 ( ) =