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Trig Identities

  1. Feb 21, 2016 #1
    I have re-post this forum as I should have paid closer attention to rules. I apologized for that.

    1. The problem statement, all variables and given/known data

    1) The expression tan^3 θ + sinθ/cosθ is equal to:

    (a) cot θ (b) tan θ sec^2 θ (c) tan θ (d) sin θ tan θ (e) tan θ csc^2 θ

    2) Simplify (cos θ/1+ sin θ - cosθ/sinθ-1)^-1

    (a) cos θ/2 (b) 2sec θ (c) 2sin θ (d) csc θ/2

    3) The expression cot θ/csc θ-sin θ is equal to:

    (a) cos θ (b) sec θ (c) tan θ (d) sin θ (e) csc

    2. Relevant equations:
    Identifying the trigonometry identities
    Cot= cos/sin
    Csc= 1/Sin

    3. The attempt at a solution:

    Here's my attempt on the first question:
    Tan^3 = Sin^3/Cos^3;
    Sin^3/Cos^3 + Sin/Cos=
    Sin^4/Cos^4 = Tan

    (c) Tan

    The second question (I'm very lost on this) :

    1/2cos= 2sec

    (b) 2Sec

    The third question:
    (Cos/Sin)/ (1/Sin) -(Sin/1) =
    (Cos/Sin)(Sin/1)/ (1/Sin)=
    Cos/(1/Sin) =
    Sin/Cos = Tan
    (c) Tan
    Last edited by a moderator: Feb 21, 2016
  2. jcsd
  3. Feb 21, 2016 #2


    Staff: Mentor

    It's not clear what the questions are.
    For 1) is it ##\tan^3(\theta) + \frac{\sin(\theta)}{\cos(\theta)}## or is it ##\frac{\tan^3(\theta) + \sin(\theta)}{\cos(\theta)}##?
    What you wrote is the first, although you might have meant the second. When you write fractions using inline text, you have to use parentheses around the entire numerator or denominator (or both), whenever there are two or more terms.

    For 2) I can't even guess what you mean.

    For 3) what you wrote is ##\frac{\cot(\theta)}{\csc(\theta)} - \sin(\theta)##. Did you mean ##\frac{\cot(\theta)}{\csc(\theta) - \sin(\theta)}##? If so, write it as cot(θ)/(csc(θ) - sin(θ)).
  4. Feb 21, 2016 #3


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    Hello Cosmic-Kat. Welcome to PF!

    You have committed two huge algebra/trig errors.

    Review how to add rational expressions (fractions).

    Sin4(θ)/Cos4(θ) = (Sin(θ)/Cos(θ))4 ≠ Tan(θ)
  5. Feb 21, 2016 #4


    Staff: Mentor

    This is the problem that Sammy said had two errors. For this problem it's better to write things in terms of ##\tan(\theta)##.
  6. Feb 21, 2016 #5
    I am sorry about the confusion.
    I should have double check my writing.
    1) The expression (tan^3 θ) + (sinθ/cosθ) is equal to

    2) Simplify ((cos θ/1)+ (sin θ - cosθ/sinθ-1))^-1

    3) The expression (cot θ)/(csc θ-sin θ) is equal to;

    Is that clear?
  7. Feb 21, 2016 #6


    Staff: Mentor

    1 and 3 are clear.
    Part of 2 is unclear: (sin θ - cosθ/sinθ-1)
    What you wrote is ##sin(\theta) - \frac{cos(\theta)}{sin(\theta)} - 1##
  8. Feb 21, 2016 #7


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    #1 and #3 are Clear enough.

    Now, for #1.
  9. Feb 21, 2016 #8
    Sorry again. I am terrible at writing!

    ((cos θ/(1+sin θ )- (cosθ/(sinθ-1))))^-1

    is this clear
  10. Feb 21, 2016 #9
    That's good to hear that #1 and #3 is better. I am not a good writer...
    And about #1?
  11. Feb 21, 2016 #10


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    Yes, What did Mark suggest for #1 ?
  12. Feb 22, 2016 #11


    Staff: Mentor

    Regarding #1, don't do little bits and pieces of it. Start with the given expression and work with it to get it in simplest form.
    (Note: I used x rather than ##\theta##, as x is easier to write.
    Don't do this: tan^3(x) = sin^3(x)/cos^3(x);
    Do this: ##\frac{sin^3(x)}{cos^3(x)} + \frac{sin(x)}{cos(x)}= tan^3(x) + ??##
    This is not equal to ##tan^4(x)##.
  13. Feb 22, 2016 #12
    I see. Do I multiply both numerators and denominators with the least common denominators?
  14. Feb 22, 2016 #13


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    Yes, if you insist on using this in "fraction" form, you will need a common denominator.

    But Mark suggested writing sin(x)/cos(x) in terms of tan(x) .

    In either case, you will eventually need to use an identity not included in the list you have under Relevant Equations ; one of the Pythagorean Identities.
  15. Mar 3, 2016 #14
    I always found this stuff hard to recall off the top of my head as well and just found it much easier to just use Euler's formulas and just start factoring out exp(i*theta) whenever possible and simplifying. For two I would just get it into some form of one function divided by another like
    then use Euler's formulas if you don't recall all the trig formulas that I can seem to remember and you should get an answer
  16. Mar 3, 2016 #15


    Staff: Mentor

    It's very likely that the OP hasn't seen Euler's fomula yet. From the problems posted, he or she is just barely getting started with trig.
  17. Mar 4, 2016 #16
    I solved it. Here are pictures:
    math1.jpg math2.jpg math3.jpg

    Also, I am physics major and I am taking trigonometry class (hopefully I'll pass).
  18. Mar 4, 2016 #17


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    That all seems to work.
    For the first one, a very useful formula is ##1+\tan^2=\sec^2##. So you can go ##\tan^3+\sin/\cos=\tan^3+\tan=\tan()(1+\tan^2)=\tan \sec^2##
  19. Mar 4, 2016 #18


    Staff: Mentor

    This is where I was heading in post #4.
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