Use this page to ask anything related to MAT 180. If you are registered and logged in, your posts will display your user name. If you would like to post anonymously, make sure you are logged out. Anyone is allowed to post.
Use this page to ask anything related to MAT 180. If you are registered and logged in, your posts will display your user name. If you would like to post anonymously, make sure you are logged out. Anyone is allowed to post.
Professor Nevo, I have 91% on Problem 16. The only part of the problem I’m unable to solve is the last one tan(u+pi/2). This is undefined, but I’m not sure how to input this into WeBWorK.
It’s not undefined. Look at the sum-and-difference formulas.
Tan Pi/2 is undefined. The problem would work out to (3/4+1/0)/(1-3/4(1/0))
Use the identity
$$
\tan (u + \pi/2) = \frac{\sin(u + \pi/2)}{\cos (u + \pi/2)}
$$
Then apply the sum formulas for cosine and sine.
Professor Nevo, I’m not sure why I got problem 13 wrong on the WeBWorK. One solution I got that worked was arccos(1/6)-3+2pi, but the same value +pi didn’t work. Is it because that value of cosine would be negative?
\(\arccos x\) returns a value between 0 and \(\pi\), inclusive. Therefore, you need to figure out the other value keeping the unit circle in mind. Then add any integer multiple of \(2\pi\) to get all solutions on the interval \((-\infty, \infty)\). Finally solve for \(x\). Range through different values of \(n\), to determine the solutions in the interval 0 to \(2\pi\), inclusive.
I’m not sure how to input the answers for problem 5 on the WeBWorK. I changed the degrees to radians but, you get decimal answers if you try to solve and WeBWork won’t take decimals.
If you want the exact value of say
$$
\cos(45^\circ)\cos(100^\circ)+\sin(45^\circ)\cos(100^\circ)
$$
Use the sum formula for cosine.
The exact value for the example given would then be
$$
\cos(45^\circ -100^\circ) = \cos(-55^\circ)
$$
The values you get for the input to the trigonometric function should be nice enough so that you can simplify without a calculator.
I have a couple of questions on homework 5 so far:
#2 We have to solve sec(v-u), and first I would change it to (1/cos(v-u)) … but being that it’s “v-u” does that change the formula we should use in the place of cosine? Because I know with addition it doesn’t matter because it’s commutative, but with subtraction it would change the value. So I’m not sure which formula I should be using now.
#3 This one confuses me because it could go two different ways… I’m not sure if the original problem translates to sin(u+v)-sin(u-v)=1/2 or sinu-sinv=1/2 ..?
For number 2, apply the formula for \(\cos(u-v)\) but with the roles of \(u\) and \(v\) reversed. That is,
$$
\cos(v-u) = \cos v \cos u + \sin v \sin u.
$$
Or, alternatively, you can use the fact that \(\cos(v-u) = \cos(-(u-v)) = \cos(u-v)\), where the last equality follows from the odd/even identities, and then use the difference formula for cosine as is.
For number 3, you’d want to think of it like the first form you listed: \(\sin(u+v) – \sin(u-v) =1/2\).
Problem 16 is causing me trouble. I’m pretty sure you have to make use of arcsin to solve the equation, so I converted the equation to arcsin(3-sin(x))/3=x. However I’m not sure what to do from there on.
You want to isolate the trigonometric function first. For example, if you wanted to solve \(7 \cos x + 9 = 14 \cos x\) for \(x\), you would want to isolate \(\cos\) on one side. You should get,
$$
7 \cos x – 14 \cos x = -9
$$
Now consolidate to get
$$
-7 \cos x = -9.
$$
Finally, divide both sides by \(-7\) to get
$$
\cos x = \frac{9}{7}.
$$
Note that there is no \(x\) such that \(\cos x = \frac{9}{7}\) since the output of the cosine function is bounded between -1 and 1.
Professor Nevo, Problem 11 and all problems similar to that are giving me trouble. To solve this one you have to make use of tan’s inverse trigonometric function, so you get arctan of some value=the angle in radians. I got tan(alpha)=-1/2, which corresponds to arctan -1/2= alpha, however there is no exact angle that would work here for the interval given. I submitted a decimal answer but it didn’t work.
Keep in mind that you don’t need to type in a decimal approximation (which should work, if correct), WeBWorK will understand
arctan(-1/2).Also keep in mind that the problem requires that you give angle values that lie on the interval \([0, 2\pi)\) and that your calculator might give you an angle outside of this interval. For example, \(\arctan(-1/2) = -0.463648\) which is not greater than or equal to zero and hence not in the interval \([0,2\pi)\). This angle lies in the fourth quadrant adding \(\pi\) to this angle puts you in the second quadrant and results in an angle greater than zero for which \(\tan\) of that angle is \(-1/2\). There is still one other value in the interval \([0,2\pi)\) that works. What is it?
Professor Nevo, how do you get a table of nice values for which to evaluate trigonometric functions? Once you find the period must you divide it by 4, and then add that to the starting value?
Yes, that’s correct. Remember that your starting value will depend on the input to the trigonometric function and the trigonometric function itself. For example, consider \(g(x) = a \sin(bx -c) + d\) then the starting value of one cycle is the \(x\) value that’s the solution to \(bx-c=0\) whilst \(h(x) = a \sec(bx-c) + d\) has a starting value \(x\) of one cycle that’s the solution to \(bx-c = -\pi/2\). Also keep in mind that the period depends on the value of \(b\) (assumed to be positive, if it is not, use the even/odd identities) and the trigonometric function.
Wait, isn’t the left asymptote determined by \(bx-c=0\) as well for \(\csc(x)\)?
Absolutely right! \(\csc(bx-c)\) has an asymptote at \(bx-c = 0\) since this is where \(\sin(bx-c)\) is equal to zero, thereby making \(\csc(bx-c)\) undefined. I had meant to use \(\sec(bx-c)\) to draw a contrast between the starting values. I have edited the response to reflect this change.
Professor Nevo, I’m having trouble with Problem 3 on the WeBWorK. The question gives you an angle of 210 degrees and the radius of its terminal side. is 12. It asks you to find the point, W, in the figure. I submitted it’s coordinates and got the wrong answer. Is the radius even necessary information?
Problem 8 is also giving me trouble. I drew the triangle in the first quadrant and labelled the sides, but you have two unknowns so you can’t solve for the third side. Does this problem require the use of Pythagorean identities?
You can use the unit circle to determine the ordered pair that corresponds to \(120^\circ\). But this will be the ordered pair corresponding to a radius of length \(1\). If the radius is scaled up, as in your problem, then you must also scale the ordered pair, accordingly.
I am having issues posting my answers for number 7 on WeBWorK. When I do sqrt, or any variation of this i am getting the answer as not allowed. How am I supposed to write out the square roots as the answer?
Yes, you have the correct solution. However, WeBWorK doesn’t understand sqrt(3). You need to make use of the fact that radicals can be written as fractional exponents. That is, Sqrt(3) can be written as 3^(1/2).
I’m having trouble with WeBWorK assignment 1, problems 17 and 19. Problem 17 gives you three pictures, and you’re being told to give the angle measure in radians. How would one go about solving. I eyeballed the answer, and was wrong.
Problem 19 has you find terminal points of P(x,y) when given P(a,b) and pi-t, -t, pi+t and 2pi+t respectively. This problem makes absolutely no sense to me.
For number 17, take a look at the graph that shows the values of angle measures of 1,2,3,… radians in section 4.1.
For number 19, the following is a description of how to do part (a).