# Math Help - Precal Test Tomorrow

1. ## Precal Test Tomorrow

Seems that others have similar problems that I do. I've finished up most of my review packet, just a few things I don't quite get yet. Wondering if someone here could clarify these two questions. Then all I have left is to face the daunting task of saving myself from my own stupidity.

1. Find the exact value of the expression. Do not use a calculator.

csc^2 70 degrees - tan^2 20 degrees

multiple choice

a) 2 b) 0 c) -1 d) 1

I think the others I am figuring out right now.

2. Originally Posted by Gerbilkit
Seems that others have similar problems that I do. I've finished up most of my review packet, just a few things I don't quite get yet. Wondering if someone here could clarify these two questions. Then all I have left is to face the daunting task of saving myself from my own stupidity.

1. Use a calculator to solve the equation on the interval 0 <= theta <= 2pi

csc^2 70 degrees - tan^2 20 degrees

multiple choice

a) 2 b) 0 c) -1 d) 1

I think the others I am figuring out right now.
you should know that $\csc^2 70 = \frac 1{\sin^2 70}$, and that you can plug in to your calculator

3. I need the exact values please. No calculators.

4. Im confused. Your question states to use a calculator making it a trivial problem but now you say you cannot use a calculator?

5. Oh erm oops. Starting typing out the wrong question. Sorry about the confusion.

6. ok, here's the trick. it might take a while for us to go through this with me giving you hints, so i'll just give you a possible solution.

things to know:

(1) $\csc x = \frac 1{\sin x}$

(2) $\sec x = \frac 1{\cos x}$

(3) $\tan^2 x = \sec^2 x - 1$

(4) the sine of an angle is equal to the cosine of its compliment (complimentary angles are angles that sum to 90 degrees).

so, for instance: $\sin 30 = \cos 60$, $\sin 45 = \cos 45$, $\sin 10 = \cos 80$, etc

particularly what we will use here is: $\cos 20 = \sin 70 \implies \cos^2 20 = \sin^2 70$

solution:

$\csc^2 70 - \tan^2 20 = \frac 1{\sin^2 70} - \left( \sec^2 20 - 1 \right)$

$= \frac 1{\sin^2 70} + 1 - \sec^2 20$

$= \frac 1{\sin^2 70} + 1 - \frac 1{\cos^2 20}$

$= \frac 1{\sin^2 70} + 1 - \frac 1{\sin^2 70}$

$= 1$

there are other solutions here. this is just the one that occurred to me first. maybe sixstringartist can come up with another one

7. Hello, Gerbilkit!

1. Find the exact value of the expression. Do not use a calculator.

. . $\csc^2\!70^o - \tan^2\!20^o$

$(a)\;2\qquad (b)\;0\qquad (c)\:-1\qquad (d)\;1$
Note the identity: . $\sin70^o \:=\:\cos20^o$

We have: . $\csc^2\!70^o - \tan^2\!20^o \;\;=\;\;\frac{1}{\sin^2\!70^o} - \frac{\sin^2\!20^o}{\cos^2\!20^o} \;\;=\;\;\frac{1}{\cos^2\!20^o} - \frac{\sin^2\!20^o}{\cos^2\!20^o}$

. . $=\;\;\frac{1-\sin^2\!20^o}{\cos^2\!20^o} \;\;=\;\;\frac{\cos^2\!20^o}{\cos^2\!20^o} \;\;=\;\;{\color{blue}\boxed{1}}$ . . . answer (d)

Too slow again . . .

8. Originally Posted by Soroban
Too slow again . . .
yes. but the identities you used were nicer. i mean who can remember that $1 + \tan^2 x = \sec^2 x$, seriously

9. Originally Posted by Jhevon
yes. but the identities you used were nicer. i mean who can remember that $1 + \tan^2 x = \sec^2 x$, seriously
Well I did but I didn't know about the complementary angle thingamajig. That certainly worked out nicely. The only concern I have is that possibly I wasn't supposed to know that, and was supposed to use a different method to solve it, well if that's the case too bad. Thanks again.

10. Originally Posted by Gerbilkit
... I didn't know about the complementary angle thingamajig... The only concern I have is that possibly I wasn't supposed to know that, and was supposed to use a different method to solve it...
i doubt you can solve this without a calculator if you didn't know that. but of course, everytime i say something like that, someone comes along and proves me wrong, so i'm waiting ...