Trig Help

1. Jan 10, 2007

Fresh4Christ

I have the problem:

0 = 3[cos(35)*cos(A) - sin(35)*sin(A)] - cos(A)
where A is alpha...my unknown degree.

somehow that turns into this:

tan(A) = [cos(35) - 1/3] / sin(35)

I am not drawing the connection or seeing how that is happening....

Could you help???? THANKS

2. Jan 10, 2007

interested_learner

Show some work. Hint: look up some trig identities and start playing around.

3. Jan 10, 2007

Fresh4Christ

i tried... this isn't a hw problem... its in the text book. It jumps from that first step to the next one just saying "Then we can see:" ... and I cant see that

4. Jan 10, 2007

tim_lou

Hint: divide everything by cos A first.

5. Jan 11, 2007

VietDao29

Ok, in fact, the book does skip some steps:
$$0 = 3[\cos (35 ^ o) \cos A - \sin(35 ^ o) - \sin A] - \cos A$$
Now divide both sides by cos A, we have:
$$0 = \frac{3[\cos (35 ^ o) \cos A - \sin(35 ^ o) - \sin A] - \cos A}{\cos A} = 3 \cos (35 ^ o) - 1 - 3 \sin (35 ^ o) \tan A$$
$$\Leftrightarrow 3 \cos (35 ^ o) - 1 = 3 \sin (35 ^ o) \tan A$$
Divide both sides by 3, we have:
$$\Leftrightarrow \cos (35 ^ o) - \frac{1}{3} = \sin (35 ^ o) \tan A$$
Now, divide everything by sin(35o), we have:
$$\Leftrightarrow \frac{\cos (35 ^ o) - \frac{1}{3}}{\sin (35 ^ o)} = \tan A$$
Can you get this? :)

Last edited: Jan 11, 2007
6. Jan 11, 2007

Fresh4Christ

Yes, thank you very much

7. Jan 11, 2007

storygeek

VietDao, how did you get your second line?

8. Jan 13, 2007

Gib Z

He divided both sides by 3, if you mean the 2nd part of that line, He just added 3 sin 35 tan A to both sides.