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

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

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

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

Hint: divide everything by cos A first.

VietDao29
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:
Yes, thank you very much

VietDao, how did you get your second line?

Gib Z
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.