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

  • #1
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
 

Answers and Replies

  • #2
Show some work. Hint: look up some trig identities and start playing around.
 
  • #3
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
682
1
Hint: divide everything by cos A first.
 
  • #5
VietDao29
Homework Helper
1,423
3
Ok, in fact, the book does skip some steps:
[tex]0 = 3[\cos (35 ^ o) \cos A - \sin(35 ^ o) - \sin A] - \cos A[/tex]
Now divide both sides by cos A, we have:
[tex]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[/tex]
[tex]\Leftrightarrow 3 \cos (35 ^ o) - 1 = 3 \sin (35 ^ o) \tan A[/tex]
Divide both sides by 3, we have:
[tex]\Leftrightarrow \cos (35 ^ o) - \frac{1}{3} = \sin (35 ^ o) \tan A[/tex]
Now, divide everything by sin(35o), we have:
[tex]\Leftrightarrow \frac{\cos (35 ^ o) - \frac{1}{3}}{\sin (35 ^ o)} = \tan A[/tex]
Can you get this? :)
 
Last edited:
  • #6
Yes, thank you very much
 
  • #7
7
0
VietDao, how did you get your second line?
 
  • #8
Gib Z
Homework Helper
3,346
6
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.
 

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