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Proving Trigonometric Identities (page 2 of 3)

  • Prove the identity sin4(x) – cos4(x) = 2sin2(x) – 1
  • I can't tell which side is more complicated, but I do see a difference of squares on the LHS, so I think I'll start there.

      sin4(x) – cos4(x) = (sin2(x) + cos2(x))(sin2(x) – cos2(x))

    The first factor, sin2(x) + cos2(x), is always equal to 1, so I can ignore it. This leaves me with:

      sin2(x) – cos2(x)

    Hmm... I'm not seeing much of anything here. But I do know, glancing back at the RHS of the identity, that I need more sines and fewer cosines. I think I'll try using the Pythagorean identity that simplified that first factor, but in a slightly different form. If sin2(x) + cos2(x) = 1, then  cos2(x) = 1 – sin2(x), and:

      sin2(x) – cos2(x) = sin2(x) – (1 – sin2(x)) = sin2(x) – 1 + sin2(x) = 2sin2(x) – 1

    And that's what I needed. For my hand-in work, I'll put it all together:

      sin4(x) – cos4(x) = (sin2(x) + cos2(x))(sin2(x) – cos2(x))

          = 1(sin2(x) – cos2(x)) = sin2(x) – cos2(x) = sin2(x) – (1 – sin2(x))

          = sin2(x) – 1 + sin2(x) = sin2(x) + sin2(x) – 1 = 2sin2(x) – 1

  • Prove the identity (1 – cos2(α))(1 + cos2(α)) = 2sin2(α) – sin4(α)
  • I think I'll start by multiplying out the LHS:

      1 – cos2(α) + cos2(α) – cos4(α) = 1 – cos4(α)

    That doesn't seem to have gotten me anywhere. What if I apply the Pythagorean identity to that first factor? Then I'll get:

      (1 – cos2(α))(1 + cos2(α)) = sin2(α)[1 + cos2(α)]

    Hmm... That doesn't seem to have helped, either. Okay, what happens if I work on the other side? I can factor a squared sine out of the two terms:

      sin2(α)[2 – sin2(α)] Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved

    If I break off a 1 from the 2, I can use that same Pythagorean identity again. (I think I'm detecting a theme....)

      sin2(α)[1 – sin2(α) + 1] = sin2(α)[1 – sin2(α) + sin2(α) + cos2(α)]

            = sin2(α)[1 + cos2(α)]

    Wait a minute! That's the same thing I ended up with on the LHS! Aha!

    While what I've done so far is not a proof, I have managed to get the two sides to meet in the middle. And sometimes that seems to be the only way to do a proof: work on the two sides until they meet in the middle, and then write something that looks like magic. I'm going to start with the LHS, work down to where the two sides meet, and then work up the RHS until I get back to the original identity:

      (1 – cos2(α))(1 + cos2(α)) = sin2(α)[1 + cos2(α)]

        = sin2(α)[1 + cos2(α) – sin2(α) + sin2(α)]

        = sin2(α)[1 – sin2(α) + sin2(α) + cos2(α)]

        = sin2(α)[1 – sin2(α) + 1]

        = sin2(α)[2 – sin2(α)]

        = 2sin2(α) – sin4(α)

This is how the textbook authors come up with those magical proofs, where you wonder how on earth they ever came up with one or another of the steps. There will be times you'll need to do this. But, for your hand-in work, make sure you do the work the right way: Work down one side to the meeting place, and then hop over to the other side and work back up to the starting place. Only by showing your steps in that way will your proof be valid.

  • Prove the identity sin2(θ)sec2(θ) + sin2(θ)csc2(θ) = sec2(θ)
  • Clearly, the LHS is the more complicated side, so I'll start there, and will convert everything to sines and cosines:

      sin^2(theta)sec^2(theta) + sin^2(theta)csc^2(theta) = sin^2(theta)/cos^2(theta) + sin^2(theta)/sin^2(theta)

    The first fraction simplifies to the tangent, and the second fraction simplifies to 1.

      sin^2(theta)/cos^2(theta) + sin^2(theta)/sin^2(theta) = tan^2(theta) + 1

    What I'm left with is one of the Pythagorean identities:

      tan2(θ) + 1 = sec2(θ)

...and that's what I needed to end up with. Putting it all together:

      all the steps, shown put together

In addition to the "working on the more complicated side" and the "converting to sines and cosines" tricks, there is one other trick that hardly ever comes up, but you might want to be familiar with it....

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Cite this article as:

Stapel, Elizabeth. "Provinig Trigonometric Identities." Purplemath. Available from Accessed


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