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Basic trigonometry concept

  1. Jul 28, 2015 #1
    When we are given a problem to find the value of say tan(120 °), I was instructed to proceed by
    tan( 90*1 +30), which lies in the second quadrant and tan is negative in the second quadrant , plus odd multiple of 90 so it becomes -cot(30). Hence tan (90+α)= -cotα. But this works only because we know that α <90.If α>=90 we could just write it as tan ( 90*2+β) and the value would change accordingly.
    Keeping this in mind, what if i am with a problem like tan2x= -cot(x+π/3). My goal is to convert the LHS to a tan function. Using tan (90+α)= -cotα (that's what my teacher did) we can write tan2x= tan(π/2+x+π/3) and continue solving. But how can we use the above method without knowing if x >90 or<90? really confused
     
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  3. Jul 28, 2015 #2

    symbolipoint

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    You can use how the angle relates to some reference angle in the first quadrant.
    120 degrees is in quadrant 2. Same sine as if 90-(120-90)=60 degrees. Cosine OPPOSITE in sign than for 60 degrees.
    Meaning: tan(120)=sin(60)/(-cos(60)).
     
  4. Jul 28, 2015 #3

    SammyS

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    (I think you meant RHS, Right Hand Side).

    It's usually not a good idea to mix radian measure and degree measure. I'll stick to radians in my reply.

    The trig function of any angle is equal to the corresponding co-function of π/2 minus that angle. This is true for any angle in any quadrant.

    In your example, that gives -cot(x + π/3) = -tan( π/2 - (x + π/3) ) .

    Tangent is an odd function so that -tan( π/2 - (x + π/3) ) = tan( -π/2 + (x + π/3) ) .

    The tangent function has a period of π, so this gives the same final result you have..

    -cot(x+π/3) = tan( π/2 + x + π/3 ) no matter what the quadrant of x .

    Added in Edit:
    Your general conjecture is also true, no matter the quadrant of α.

    tan(π/2 + α) = cot(π/2 - (π/2 + α)) = cot(-α) = -cot(α) .
     
    Last edited: Jul 28, 2015
  5. Jul 28, 2015 #4
    If suppose x+π/3 lies in the first quadrant, then adding π/2 to it will make cot negative and the RHS would become positive.
    But if x+π/3 was lying in another quadrant like the second quadrant, then adding π/2 to it will make cot positive and the RHS would remain negative which is what I don't want. Doesn't the value of x matter?
     
  6. Jul 28, 2015 #5

    symbolipoint

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    Certainly the value of x is important. In your example of the first post, you have 120 degrees, or [itex]pi/2+pi/3[/itex]. Use the Unit Circle to cut through the worded discussion and your intuition should help in analyzing that. You can then focus on the sines and cosines, and decide which reference angle in quadrant 1 to use.
     
  7. Jul 28, 2015 #6

    SammyS

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    In each quadrant, tan and cot have the same sign.

    As you add π/2 , you go to the next higher quadrant.

    The tan & cot change sign as you rotate through the quadrants.
     
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