Review Problems for Test 2

Transcription

1 Review Problems for Test Math These problems are meant to help you study. The presence of a problem on this sheet does not imply that there will be a similar problem on the test. And the absence of a problem from this sheet does not imply that the test does not have such a problem.. a) Find a function of the form y A + B sinkx c) which has the graph shown below. π/ π - b) Find a function of the form y A + B coskx c) which has the graph shown below. y.. x -4. a) Find the exact value in degrees of sin ). b) Find the exact value in radians of tan ).. Suppose tan a b θ. Find cosθ. 4. a) tan x cotx tanx) sinx. ) ) b) +. cotx tan x c) + π ). 5. a) Compute the exact value of sin 5. b) Compute the exact value of cos π 8.

2 . a) Find all solutions to sinx) + sinx 0 in exact degrees. b) Find all solutions to in exact radians, for 0 x < π. 7. a) Find all solutions to sinx) 0.4) in radians to at least 5 decimal places. b) Find all solutions to tan x 0.4 in degrees to at least 5 decimal places, for 0 x < 0. Solutions to the Review Problems for Test. a) Find a function of the form y A + B sinkx c) which has the graph shown below. π/ π - The amplitude is, and there is no vertical translation. The graph goes through one-quarter of a period from π to π, a length of π π π units. Therefore, one period is 4 π π units. The multiplier for a period π sine curve is k π π. A normal period π sine curve starts a period at x 0. The curve in the picture starts a period at x π. Therefore, the given curve has a phase shift of π units to the right. y sin x π ) is a function which produces the graph. b) Find a function of the form y A + B coskx c) which has the graph shown below. y.. x -4

3 4) The max is at y and the min is at y 4. Therefore, the amplitude is.5. An untranslated cosine curve with amplitude.5 would have a min at y.5. The min for this curve is at 4; hence, the vertical translation is 8.5 units upward. There is a min at x. and a max at x.. The distance between the two is.. ; this is half a period, so a whole period is 4. The multiplier in the equation is π 4 π. An unshifted cosine curve has a max at x 0; the given curve has a max at x.. Therefore, the phase shift is. units to the right. y cos π x.) is a function which produces the graph.. a) Find the exact value in degrees of sin ). sin ) 90. b) Find the exact value in radians of tan ). tan ) π.. Suppose tan a b θ. Find cosθ. tan a b θ means that tan θ a b. a + b a From the triangle, I see that cosθ θ b a + b. b 4. a) tan x cotx tanx) sinx. tan x cotx sinx sinx cos x ) ) sinx tanx) sinx. )

4 b) ) ) +. cotx tan x ) ) cotx tan x ) ) sinx sinx sinx ) ) cos x sinx)cos x + ) +. c) + π ). Using the angle addition formula for sines, I have + π ) cos π + sin π sinx 0 + ). 5. a) Compute the exact value of sin 5. Using the angle addition formula for sines, I have sin 5 sin0 +45 ) sin 0 cos45 +sin45 cos0 ) ) ) + )+. b) Compute the exact value of cos π 8. Using the identity cosθ) + cosθ), I have cos π ) + cos π ) + ) Since π 8 is in the first quadrant, cos π 8 is positive. Hence, cos π a) Find all solutions to sinx) + sinx 0 in exact degrees. ) + sinx 0, ) + ) 0, or. has no solutions, since for all x. has the solution x 70. Therefore, the solutions are x 70 + n 0, where n is an integer. 4

5 b) Find all solutions to in exact radians, for 0 x < π., 0, ) 0. 0 has the solutions x π and x π. 0 gives. To find solutions to this equation, I first find the angles between 0 and π which have cosine equal to. π/ 5 π/ - Thus, cos π cos 5π I get more solutions by adding π to these solutions. Since the original equation had x instead of x, I need solutions in consecutive periods: cos 7π cos π cos π cos 7π Now I set x equal to each of the angles above, and solve each equation for x: x π, x π x 5π, x 5π x 7π, x 7π x π, x π x π, x π x 7π, x 7π 5

6 The solutions are x π, x π, x π, x 5π, x 7π, x π, x π 7π, and x. 7. a) Find all solutions to sinx) 0.4) in radians to at least 5 decimal places. ) 0.4), ±0.4. This gives one solution in each quadrant: π/ 5 π/ - I find θ using the inverse sine function. I can t find all the solutions this way, since sin ) only produces angles between π and π.) I get θ sin π θ π sin π + θ π + sin π θ π sin The solutions are x πn, x.0 + πn, x πn, and x πn, where n is an integer. b) Find all solutions to tan x 0.4 in degrees to at least 5 decimal places, for 0 x < 0. There is one solution between π and π : π/ 5 π/ - I have θ tan

7 I get additional solutions by adding 80 to the first solution since the tangent function is periodic with period 80 ). Since the original equation had x instead of x, I need solutions in consecutive periods. Thus, tan tan I set x equal to each of these two angles, and solve for x: x.804, x x 0.804, x The solutions are x n 80 and x n 80, where n is an integer. A lot of people my age are dead at the present time. - Casey Stengel c 009 by Bruce Ikenaga 7