#$&* course Mth 164 3/6 5:30 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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**** Query problem 6.5.10 exact value of sin^-1( -`sqrt(3)/2)
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sin^-1( -`sqrt(3)/2) = the angle (theta) within the interval [-pi/2, pi/2] whose sine is -`sqrt(3)/2, therefore : sin^-1( -`sqrt(3)/2) = theta = -pi/3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The inverse sine is between -`pi/2 and `pi/2. We are looking for the angle between these limits whose sine is -`sqrt(3) / 2. Since at this point we know how to construct the unit circle with values of the x and y coordinates for all angles which are multiples of `pi/4 or `pi/6, we know that the angle 5 `pi / 6 gives us y coordinate -`sqrt(3) / 2. This angle is coterminal with an angle of -`pi/3, so sin^-1(-`sqrt(3) / 2) = -`pi / 3. ** INCORRECT STUDENT SOLUTION Using the unit circle we can find that sin(sqrt(3)/2) is pi/3. The negative sign in front of the equation will make the pi/3 negative giving us -pi/3 as the value. INSTRUCTOR RESPONSE AND CORRECTION The negative sign means that we are looking for a negative value of the y coordinate, in the unit-circle model. This doesn't automatically make the angle negative. It works out that way for the inverse sine function, but would not for the inverse cosine. The correct relationship would be sin(-pi/3) = - sqrt((3) / 2). It would be correct to say that sin^-1(pi/3) = sqrt(3) / 2, but not that sin(sqrt(3)/2) = pi / 3.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 Query problem 6.5.34 exact value of cos(sin^-1(-`sqrt(3) / 2 )?
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20:24:59 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: cos(sin^-1(-`sqrt(3)/2) We let theta = sin^-1(-`sqrt(3)/2). Then sin(theta) = -`sqrt(3)/2, where -pi/2 <= theta <= pi/2. We seek cos(theta). Because sin(theta) < 0, it follows that theta lies in quadrant IV. Since sin(theta) = -`sqrt(3)/2 = y/r, we let y = -`sqrt(3) and r = 2. The point (x, -`sqrt(3)/2), x > 0, is on a circle of radius 2, x^2 + y^2 = 4. Then x^2 + (-`sqrt(3))^2 = 4 x^2 + 3 = 4 x^2 = 1 x = 1 Then we have x = 1, y = -`sqrt(3), and r = 2 so cos(sin^-1(-`sqrt(3)/2) = cos(theta) = x/r = 1/2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** sin^-1(sqrt(3) / 2) is the angle whose sine is sqrt(3) / 2. It is therefore defined by a triangle whose base angle is opposite a 'downward vertical' side of length sqrt(3) and hypotenuse of length 2. This triangle has adjacent side sqrt ( c^2 - a^2 ) = sqrt( 2^2 - (sqrt(3))^2) = sqrt(4 - 3) = sqrt(1) = 1. The cosine of the base angle is therefore cos( sin^-1(-sqrt(3)/2)) = adjacent side / hypotenuse = 1 / sqrt(3) = sqrt(3)/3. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ???I have obtained a different solution. ???The hypotenuse length of the triangle is 2. ------------------------------------------------ Self-critique Rating: 2 **** Query problem 6.5.88 cos(tan^-1(v)) = 1 / `sqrt(1+v^2) explain how you establish the given identity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To establish the identity: cos(tan^-1(v)) = 1 / `sqrt(1+v^2) we let theta = tan^-1(v). Then tan(theta) = v, where theta is in interval (-pi/2, pi/2), and -infinity < v < infinity. We seek cos(theta). Since tan(theta) > 0, we know that theta lies in quadrant I. So we have as follows: 1 / `sqrt(1+v^2) = 1 / `sqrt(1+(tan(theta))^2) = 1 / sec(theta) = cos(theta) = cos(tan^-1(v)) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** tan(`theta) = v defines a right triangle with side v opposite the angle `theta and side 1 adjacent to `theta. The hypotenuse would thus be `sqrt(1+v^2). It follows that cos(`theta) = adjacent side / hypotenuse = 1 / `sqrt(1+v^2). Thus cos(tan^-1(v)) = adjacent side / hypotenuse = 1 / `sqrt(1+v^2).** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After sketching the triangle, I think that I understand this approach. tan(theta) = y/x = opposite side / adjacent side = v/1 = v cos(theta) = x/r = adjacent side / hypotenuse = 1/`sqrt(1+v^2) sin(theta) = y/r = opposite side / hypotenuse = v/`sqrt(1+v^2) I am not yet accustomed to visualizing the trigonometric functions as relation of triangle sides. ------------------------------------------------ Self-critique Rating: 3