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 homework problem... its in the textbook. It jumps from that first step to the next one just saying "Then we can see:" ... and I can't see that

 

[tim_lou]

Hint: divide everything by cos A first.

 

[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(35^o), we have:
\Leftrightarrow \frac{\cos (35 ^ o) - \frac{1}{3}}{\sin (35 ^ o)} = \tan A
Can you get this? :)

 

[Fresh4Christ]

Yes, thank you very much

 

[storygeek]

VietDao, how did you get your second line?

 

[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.