Section 7 (Inverse Trig Functions)

Nov 6th, 2010 by corricelli

Hello Teams 2 and 7,

Let’s use this spot to post questions for Honors Precalculus, Unit 4, Section 7!

Happy blogging,
Mrs. Corricelli

10 Responses to “Section 7 (Inverse Trig Functions)”

Jeff Murray says:

November 23, 2010 at 5:58 pm

For number 41 on tonight’s homework, I found the arctan of all three coordinates as a means of getting the other coordinate, but when I looked at the back at the book and on the Calcchat website, it only finds the arc tan for the first coordinate (-root3, ___) and then they switch to finding the tangent of the last two coordinates. I’m completely clueless as to why you do that. thanks

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Colleen Lynch says:

November 23, 2010 at 7:15 pm

I noticed the same thing but have no idea to help you because I just as lost

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corricelli says:

November 23, 2010 at 7:43 pm

Jeff and Colleen:

With 41, they are either giving you y values, like pi/4 or -pi/6 or they are giving you x values, like -root(3).

So, lets look at pi/4. If y=arctan(x)=pi/4, then this means we are in a situation like we discussed in class.
arctan(x)=pi/4 means that tan(pi/4)=x
So, x=tan(pi/4) = 1.

Similarly, when arctan(x)=-pi/6, this means:
arctan(x)=-pi/6 and also that tan(-pi/6)=x
So, x=tan(-30degrees)=-root(3)/3.

Alternatively, when x=-root(3), this means that we are looking for y. y=arctan(-root(3)). -root(3)=(-root(3)/2)/(1/2). So the angle, y is opposite -root(3)/2 in the restricted region from -pi/2 to pi/2, not including either. So this means we are in the fourth quadrant and opposite -root(3)/2 is -60 degrees or -pi/3.

Also, please note that each of these solutions is written out after the notes in the packet given in class.

Thank you,
Mrs. Corricelli

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Taylor Schumann says:

November 23, 2010 at 10:06 pm

For #45, we just had to find a function, which turned out to be:

inverse sin ((x+2)/5) = theta

I was wondering if we were ever going to have to solve for theta, or do we need an x value to do this? Or maybe, solving the equation do we need a value for theta?

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Will Thomas says:

November 23, 2010 at 11:07 pm

for anyone that is at all confused, i would check out this video, http://www.youtube.com/watch?v=LLbOzKodfbY&feature=channel. there are full lessons on the website by professor ed, for about $2 each. the website is mindbites.com, and i’ve purchased one and it’s been really helpful.

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Sabrina Jin-Corcoran says:

November 28, 2010 at 12:30 pm

I don’t know if this is the right place but…for #2b on the study guide I think the answer is wrong.
The question was: csc(-60°) + sin(-45°) + sin(90°) × cot(-270°)
For this problem, I first solved sin(90°) × cot(-270°) which equaled 0.
cot(-270°)=(0÷-1)=0 and sin(90°)=(1) so when multiplied together they = 0.
(the study guide answers said that sin(90°) + cot(-270°) = 1+(0÷-1)=1)
The rest of my work was the same as the study guide so csc(-60°) was = to ((-2 )÷3) and
Sin (-45°) was = to (-√2)÷2
So the answer should be (-4√3 – 3√2)÷6
Not ((-4√3 – 3√2 +6)÷6)
Just for anyone who might have been confused.

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Michaela Celella (per 6) says:

November 28, 2010 at 4:24 pm

i think youre right sabrina!
but, did anyone find something funky with 2a also? it looked like the values for cotangent might have been changed

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Mariah Castillo says:

November 28, 2010 at 5:33 pm

yea, in the study guide for 2b, it’s telling you to add instead of multiply.

bu i didn’t see anything wrong for 2a. cot is cos/ sin. the only qualm i had with that is that sin (-3pi/2)= 1, not -1. it doesn’t change the over all answer, though.

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Ryan Phelan says:

November 28, 2010 at 7:53 pm

hey guys on the study guide for number 8a) how come you can just make it equal to alpha? what makes that possible? are you just supposed to automatically do that and time it has the -1?

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corricelli says:

November 28, 2010 at 8:17 pm

Hello,

Three things:
1.) I know that I made that mistake and that sin (-3pi/2)= 1. I announced this in class when I handed this out to help you to get ready for your test. The original problem was sin (3pi/2). I was not worried about the change because I knew that 0/1 = 0/-1 = 0, but I did forget when I did the problems out for you.
2.) These comments should be under the unit review, not section 7.
3.) Ryan, you can pick any letter. I knew the output for an inverse function is an angle so Greek letter alpha made sense to me for part a. The point is not the letter but the way it is used.

Take care,
Mrs. Corricelli

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