#$&* course mth 164 5pm 10/18 SOLUTIONS/COMMENTARY FOR QUERY 6......!!!!!!!!...................................
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I am trying but just do not understand how to solve this problem, I’ve looked in the book but it offers no guidance for me. Where is there a -pi/3 on the unit circle.
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20:24:59 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** sin^-1(sqrt(3) / 2) is the angle whose sine is sqrt(3) / 2. It is therefore defined by a triangle whose base angle is opposite a 'downward vertical' side of length sqrt(3) and hypotenuse of length 2. This triangle has adjacent side sqrt ( c^2 - a^2 ) = sqrt( 2^2 - (sqrt(3))^2) = sqrt(4 - 3) = sqrt(1) = 1. The cosine of the base angle is therefore cos( sin^-1(-sqrt(3)/2)) = adjacent side / hypotenuse = 1 / sqrt(3) = sqrt(3)/3. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have the same issue as before. ------------------------------------------------ Self-critique Rating:2 **** Query problem 6.5.88 cos(tan^-1(v)) = 1 / `sqrt(1+v^2) explain how you establish the given identity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I can do alright on the qa part of these assignments but I get confused on these assignments. ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: