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Thread: Proving trig identities: cos(g+h)cos(g-h) = cos^2(g)-cos^2(h)

  1. #1

    Proving trig identities: cos(g+h)cos(g-h) = cos^2(g)-cos^2(h)

    Hello everyone! I want to thank everyone who helps me in advance as I'm seriously struggling with two problems. The first one I think I managed to solve, but I would like to know for certain (I will post all the work I did for it) while the second question has me completely confused. The instructions for the questions is to simply verify the trigonometric identities.

    2) cos(g+h)cos(g-h) = cos2(g)-cos2(h)

    I decided to manipulate the right side and here is what I did:

    LHS (left-hand side) = (cos(g)-cos(h))2
    LHS = (cos(g)+cos(h))(cos(g)-cos(h))
    LHS = cos(g+h)cos(g-h)
    cos(g+h)cos(g-h) = cos(g+h)cos(g-h)

    That is my final answer, but I would like to know whether I managed to solve the problem properly or if I made a logical error somewhere.

    3) (sin(x)-tan(x))(cos(x)-cot(x)) = (cos(x)-1)(cos(x)-1)

    As for this problem, I've tried manipulating both sides (not at the same time), and have gotten stuck. Currently, my work is as follows:

    (sin(x)-(sin(x)/cos(x)))(cos(x)-(cos(x)/sin(x))) = RHS (right-hand side)

    After this part is where I get confuse. Should I subtract the sin(x) and cos(x) in the numerator? I would end up with this:

    (1/cos(x))(1/sin(x)) = RHS

    Which will give me this:

    (sec(x))(csc(x)) = RHS

    And that's as far as I managed to get. Any help is appreciated!

  2. #2
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    Quote Originally Posted by rojareina View Post
    2) cos(g+h)cos(g-h) = cos2(g)-cos2(h)

    I decided to manipulate the right side and here is what I did:

    LHS (left-hand side) = (cos(g)-cos(h))2
    LHS = (cos(g)+cos(h))(cos(g)-cos(h))
    LHS = cos(g+h)cos(g-h)
    cos(g+h)cos(g-h) = cos(g+h)cos(g-h)

    That is my final answer, but I would like to know whether I managed to solve the problem properly or if I made a logical error somewhere.
    As far as I can tell, exactly nothing you've done here is correct. For one thing, I'm very confused because you say "I decided to manipulate the right side" but then begin by saying that the left hand side is equal to something. I can explain that away as a typo somewhere, although I can't be certain where. However, neither the left-hand nor the right-hand side is equal to [tex][\cos(g) - \cos(h)]^2[/tex].

    If we let [tex]\alpha = \cos(g)[/tex] and [tex]\beta = \cos(h)[/tex], it should become clear why the right-hand side is not equal to this, because [tex]\alpha^2 - \beta^2 \ne (\alpha - \beta)^2[/tex]. And using the angle addition and subtraction formulas reveal that the left-hand side is not equal to this either.

    [tex]\cos(g+h) \cos(g-h) = [\cos(g) \cos(h) - \sin(g) \sin(h)][\cos(g) \cos(h) + \sin(g) \sin(h)] = \cos^2(g) \cos^2(h) - \sin^2(g) \sin^2(h)[/tex]

    Quote Originally Posted by rojareina View Post
    3) (sin(x)-tan(x))(cos(x)-cot(x)) = (cos(x)-1)(cos(x)-1)

    As for this problem, I've tried manipulating both sides (not at the same time), and have gotten stuck. Currently, my work is as follows:

    (sin(x)-(sin(x)/cos(x)))(cos(x)-(cos(x)/sin(x))) = RHS (right-hand side)

    After this part is where I get confuse. Should I subtract the sin(x) and cos(x) in the numerator? I would end up with this:

    (1/cos(x))(1/sin(x)) = RHS

    Which will give me this:

    (sec(x))(csc(x)) = RHS

    And that's as far as I managed to get. Any help is appreciated!
    This, too, seems to be plagued by strange notation, wherein you say you're manipulating the right-hand side, but then actually manipulate the left-hand side. I'm also guessing there was another typo and you actually meant the right-hand side to be [cos(x) - 1][sin(x) - 1]. That change makes it actually be an identity. In any case, your first step is good, but after that it's not correct due to math errors. I can't see how you got from:

    [tex]\left[ \sin(x) - \dfrac{\sin(x)}{\cos(x)} \right] \left[\cos(x) - \dfrac{\cos(x)}{\sin(x)} \right][/tex]

    To:

    [tex]\dfrac{1}{\cos(x)} \cdot \dfrac{1}{\sin(x)}[/tex]

    I assume you tried to follow the usual steps of multiplying by a form of 1 (in this case [tex]\dfrac{\cos(x)}{\cos(x)}[/tex]) to get a common denominator... but then what?

  3. #3
    Quote Originally Posted by ksdhart2 View Post
    As far as I can tell, exactly nothing you've done here is correct. For one thing, I'm very confused because you say "I decided to manipulate the right side" but then begin by saying that the left hand side is equal to something. I can explain that away as a typo somewhere, although I can't be certain where. However, neither the left-hand nor the right-hand side is equal to [tex][\cos(g) - \cos(h)]^2[/tex].

    If we let [tex]\alpha = \cos(g)[/tex] and [tex]\beta = \cos(h)[/tex], it should become clear why the right-hand side is not equal to this, because [tex]\alpha^2 - \beta^2 \ne (\alpha - \beta)^2[/tex]. And using the angle addition and subtraction formulas reveal that the left-hand side is not equal to this either.

    [tex]\cos(g+h) \cos(g-h) = [\cos(g) \cos(h) - \sin(g) \sin(h)][\cos(g) \cos(h) + \sin(g) \sin(h)] = \cos^2(g) \cos^2(h) - \sin^2(g) \sin^2(h)[/tex]



    This, too, seems to be plagued by strange notation, wherein you say you're manipulating the right-hand side, but then actually manipulate the left-hand side. I'm also guessing there was another typo and you actually meant the right-hand side to be [cos(x) - 1][sin(x) - 1]. That change makes it actually be an identity. In any case, your first step is good, but after that it's not correct due to math errors. I can't see how you got from:

    [tex]\left[ \sin(x) - \dfrac{\sin(x)}{\cos(x)} \right] \left[\cos(x) - \dfrac{\cos(x)}{\sin(x)} \right][/tex]

    To:

    [tex]\dfrac{1}{\cos(x)} \cdot \dfrac{1}{\sin(x)}[/tex]

    I assume you tried to follow the usual steps of multiplying by a form of 1 (in this case [tex]\dfrac{\cos(x)}{\cos(x)}[/tex]) to get a common denominator... but then what?
    In response to your first question, about me saying I manipulatedthe right side, I honestly do not understand how I worded that strangely. I’mjust doing what my professor did during class – she would always manipulate oneside by setting the other side equal to it. This is the only way I know how towork trig identity problems. As for the rest, these two problems were given aspractice for an exam and I honestly did not think my professor would give usproblems that could not be proven.
    As for the second problem, there was no typo. The equation Ipreviously wrote is exactly the equation that was given to me. Does this meanthis problem can’t be proven either? I'm just so confused…my professor wants usto understand how to prove identities and yet it seems she gave my class twoproblems that can’t be proven.


  4. #4
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    Quote Originally Posted by rojareina View Post
    In response to your first question, about me saying I manipulatedthe right side, I honestly do not understand how I worded that strangely …
    Using 'LHS' as a replacement for writing out the left-hand side of an equation is not something all instructors do. Some readers could be confused by this, particularly because the right-hand side in your second line does not follow from the first.

    … I honestly did not think my professor would give usproblems that could not be proven.
    Not intentionally, of course. As ksdhart mentioned, there could be typos (in the materials that you received or in the materials that the instructor used).

    As for the second problem, there was no typo. The equation Ipreviously wrote is exactly the equation that was given to me. Does this meanthis problem can’t be proven either?
    Yes, it is not an identity. In each exercise, you can use a scientific calculator to evaluate each side for arbitrary values in the domain of each side.

    For example, in the second exercise, substituting Pi/3 for x leads to (rounded):

    0.06699 = 0.25

    If you google for trig identity practice exercises, I'm sure you'll find many at different levels of difficulty.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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