# Trig 0 = 3[cos(35)*cos(A) - sin(35)*sin(A)] - cos(A)

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

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

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

tim_lou
Hint: divide everything by cos A first.

Homework Helper
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:
Fresh4Christ
Yes, thank you very much

storygeek
VietDao, how did you get your second line?

Homework Helper
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.