Trig test review [closed]

Well I just failed a trig test.

Any help on why I did things wrong or what went wrong in my thought process? I triple checked all my answers and was positive I had 100 percent on this test, instead I got like 30 percent.

1) Show that $\tan^2 \theta - \sin^2 \theta = \tan^2 \theta \sin^2 \theta$

I turned the left side into $(\sin\sin)/\cos\sin$ which cancels out into $\sin^2/\cos^2 \sin^2$ whichs is $\tan^2 \sin^2$

2) I was suppose to "evaluate" (no idea what that means) cos 105 degrees and I got $-\sin 15$ degrees, which is wrong. I am supose to find the value of that but I have no idea how without a calculator.

4) evaluate $\tan a/2$ if $\sin^2a = .64$ and $a$ is in quadrant II I got $\tan = 1$ but in my work I forgot I negative sign and got the whole problem wrong.

5) show that $\sin3a = \sin3a - 3\sin - 4 \sin ^3 a$

I know that $3\sin a = \sin^3a$ so I reduced that to $a$ and I know $A = A - 4\sin^3$ is impossible, no number can equal itself minus 4 times something besdies 0 and $\sin$ can't equal zero for that value.

6) evaluate ($\tan83$ degrees - $\tan 38$ degrees)/ ($1+\tan 83$ degrees $\tan38$ degrees) I know that is equal to $\tan(A-B)$ and from that I got $8/8 - 1/8$ somehow equaling $7/8$ not sure where I went wrong here.

8) express $\cot 2a$ in terms of $\cot a$

I made this simple $\cot a + \cot a2a / 2$

Not sure why that is wrong, but I got all the points off.

9) simplify $(1+\tan^2\theta)/(1+\cot^2\theta)$ I made that $\sec^2/\csc^2$ which turns into $\sec^4$ I got this wrong somehow.

I always mess up tests like this for some reason, not sure what to do to pass classes. I still mess up simple math, was looking at someone's homework for math 70 and I couldn't do most of it. I probably need to drop out of this class since I am failing highschool math in college, and go back to math 70 or whatever.

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closed as too localized by Grigory M, Andres Caicedo, yunone, Asaf Karagila, t.b.Jun 29 '11 at 0:28

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Why use offensive language in a question on a public site? That is not polite at all. I removed the phrase. Please do not use that language again. –  Beni Bogosel Jun 28 '11 at 20:53
I assumed everyone here is an adult, just what I am used to. –  Adam Jun 28 '11 at 20:54
If all of us are adults, then it doesn't mean that you can speak dirty. You are asking us for help, talking to a various audience, from teachers, students, non-professionals. Is that how you speak with your teachers? Your irritation can be expressed in terms which are not offending, and better off not be expressed at all. –  Beni Bogosel Jun 28 '11 at 21:02
@Adam, as an adult, you should learn to be respectful of others. In this case, there are probably people who prefer not to see such language. Indeed, I am one of them. –  The Chaz 2.0 Jun 28 '11 at 21:15
Adam: it would be good if you'd upvote any number of answers...all said, I think all your questions have been answered in one or more of the answers posted below. People are taking time to try and help/explain, so you can better understand, and it is simply a matter of being polite by upvoting answers which are helpful... –  amWhy Jun 28 '11 at 22:56

1) $\tan^2 \theta-\sin^2 \theta = \sin^2\theta (\frac{1}{\cos^2\theta}-1)=\sin^2\theta \frac{1-\cos^2\theta}{\cos^2 \theta}$. From here on you use that $\sin^2\theta+\cos^2\theta=1$.

2) $\cos(105)=\cos(90+15)=\cos 90\cos 15-\sin 90\sin 15=-\sin 15$

5) $\sin(a+2a)=\sin a \cos 2a -\cos a \sin 2a$. Next, use formulas for $\sin,\cos$ of the double angle and the fundamental relation $\sin^2+\cos^2=1$.

6) I don't know why you missed the answer, since you wrote the correct formula $\tan(a-b)=\tan(83-38)=\tan 45$.

For you not to mess up next time, you should know well your formulas, do not hurry when doing calculations, and do not assume that the problems are harder than your level. These are basic trigonometry, so careful use of the formulas would give you the result in two lines of text at most.

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Thanks. These might be basic but I really struggle with basic math. I tried to memorize the formulas but there was just too many I couldn't remember them. I did all the homework from the book and I still couldn't recall any of them except the four cos(A+B) set. –  Adam Jun 28 '11 at 21:05
You should memorize them by exercise. Ten exercises which use $\sin(a\pm b)$ will guarantee you not to forget the formula for a while. That can be said for $\cos(a\pm b),\tan(a\pm b)$. Try and learn them in a logical manner, learn from where they come. If you do not do that, you will keep forgetting them and trigonometry will be a nightmare. So take your textbook and do all the exercises. :) –  Beni Bogosel Jun 28 '11 at 21:08
I am basically done with this class, I will likely pass with a low C. I am taking calculus next, what do I need to know for calculus? I don't care about my trig grade, no matter how hard I try the grade will not change. –  Adam Jun 28 '11 at 21:11
For calculus, you will need trig formulas. If you don't learn the formulas now, you'll have problems with them again. –  Beni Bogosel Jun 28 '11 at 21:14
All 21 or so formulas? –  Adam Jun 28 '11 at 21:15

8) Express $\cot(2a)$ in terms of $\cot(a)$. I made this simple $\cot a + \cot a2a / 2$.

I'm not clear about what you mean by $\cot a + \cot a2a / 2$, particularly the $a2a$ argument of the second $\cot$.

If you know $\tan(2a)$, or even $\tan(a + b)$ (where in this case you'd use $\tan(a + a)$, then you can figure out $\cot(2a)$, or $\cot(a + a).$

You can also proceed by remembering $$\tan(2a) = \frac{\sin(2a)}{\cos(2a)},\ \ \text{so}\ \ \cot(2a) = 1/\tan(2a) = \frac{\cos(2a)}{\sin(2a)},$$ and then use double angle formulas for $\cos(2a)$ and $\sin(2a)$, with the aim of ending with a function expressed in terms of $\cot(a) = \cos(a)/\sin(a)$.

The key thing is to be clear about the "starting point" (left hand side), with your eyes to the desired "ending point" (right hand side), whenever trying to establish an identity. This will rule out the use of some identities, or point to particularly helpful identities. So, for example in this problem, however you chose to proceed (e.g. if you started by converting $\cot(2a)$ to $\cos(2a)/\sin(2a)$, because you were more comfortable with double angle formulas) keep in mind that you are aiming to express $\cot(2a)$ in terms of $\cot(a)$ (i.e., where the angle $a$ is the sole argument, and the only trig function involved is $\cot(a)$, (perhaps added to a number, or multiplied by a factor, or raised to a power...etc.

I'll just point out that $$\tan(2a) = \frac {2\tan (a)}{1 - \tan^2(a)},$$ and since $\cot (2a) = 1/\tan (2a),$ we have $$\cot (2a) = \frac{\cot^2(a) - 1}{2\cot(a)}$$

At any rate: keep in mind that there are usually a number of ways to proceed (as I mention, there are a number of ways to express $\cot(2a)$. It helps to know the various "tools" (identities) that you can use, but to select one strategically (with the aim in mind of where you want to end up). The more you play around with the various trig identities, convincing yourself of the equivalences, and challenging yourself to derive them, the better you will master them/remember them. Memorization only works for periods of time, if you don't understand why the identities hold, and find ways to verify an identity quickly, in case your recall is shaky.

In short, your goal should be understanding the identities, which will facilitate recall, and flexibility in adapting an identity to a particular problem, rather than trying to force memorization, which works in a "rigid" manner: usually working only when problems are in the precise format of the identities you memorized.

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simplify $\dfrac{(1+\tan^2\theta)}{(1+\cot^2\theta)}$ I made that $\dfrac{\sec^2\theta}{\csc^2\theta}$ which turns into $\sec^4\theta$ I got this wrong somehow.

You were really close.

$\sec^2\theta = \dfrac{1}{\cos^2\theta}$

$\dfrac{1}{\csc^2\theta} = \sin^2\theta$

Multiply those, and you get $\dfrac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta$

I can't realistically do all these problems(they take too long to write!), but the important thing is not to give up and ask specific questions of specific steps that you don't know. You had things right until that last step.

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I am more concerned with learning why I did things wrong even though I checked them over repeatedly and was positive I was right. Any answers that are posted are of course highly appreciated though. –  Adam Jun 28 '11 at 20:51

$\cos 105^\circ$ is indeed the same as $-\sin 15^\circ$. And 15 is half of 30. So if you know the sine and cosine of $30^\circ$, you can use half-angle formulas.

If you know that $\sin(x+y) = \sin x \cos y + \cos x \sin y$, and you let $x$ and $y$ both be $15^\circ$, then on the left side you've got $\sin 15^\circ$. That tells you what $2\sin15^\circ\cos 15^\circ$ is. With that and a similar identity for the cosine of a sum, you should also be able to deduce something.

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That's a lot of deducing? How about cos(45 + 60) = ... ? –  The Chaz 2.0 Jun 28 '11 at 21:16

for question 2: $\cos(105) = \cos(60+45) = \cos60 \cos 45 - \sin 60 \sin 45 = \frac{1}{2} \cdot \frac{\sqrt {2}} {2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt 2}{2} = \frac{\sqrt 2}{4}\cdot (1-\sqrt {3})$

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