Assignment 6 query

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course Mth 164

SOLUTIONS/COMMENTARY FOR QUERY 6......!!!!!!!!...................................

 

  ****   Query   problem 6.5.10 exact value of sin^-1( -`sqrt(3)/2)

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Your solution:  

 sin^-1(-sqrt(3)/2)

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.

 

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.

confidence rating #$&*2

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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. **

 

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Self-critique (if necessary):

 I got the right answer but I’m not sure if I went about it the same way as the given solution did.

 

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Self-critique rating #$&*Ok

 

 Query   problem 6.5.34 exact value of cos(sin^-1(-`sqrt(3) / 2 )?

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Your solution:

 

 cos(sin^-1(-sqrt(3)/2)=

cos(-pi/3)=

 1/2

confidence rating #$&*2

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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 / 2. **

 

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Self-critique (if necessary):

 I just used the answer I got before in the first one question and substituted in but it did not come out to what the answer was and I am not quite sure why it did not workout that way.

The solution 1/2 is correct. In the very last line of the given solution I used sqrt(3) for the hypotenuse, resulting in an incorrect answer. The hypotenuse of the triangle in the given solution is clearly 2, and the correct answer is 1/2.

 

 

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Self-critique rating #$&*

  ****   Query   problem 6.5.88  cos(tan^-1(v)) = 1 / `sqrt(1+v^2) 

explain how you establish the given identity. 

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Your solution:  

 

 Using a triangle to find the sides we can see that because of the tan we get a hypothenuse that is equal to sqrt(1+v^2). In turn we can get that the cos is equal to the adjacent side over the hypotenuse and that leaves us with 1/sqrt(1+v^2). 

 

 

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).**

 

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Self-critique (if necessary):

 

 

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Self-critique rating #$&*Ok"

&#Your work looks good. See my notes. Let me know if you have any questions. &#

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