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Trig Identities Cheatsheet

First, let's remember our negative angle rules (N for Negative), since it turns out you can always use these rules to derive any of the identities below from any of the other ones!

(N1: Negative Angle Negates

sin)

sin(θ)=sin(θ)

(N2: Negative Angle Doesn't Change

cos)

cos(θ)=cos(θ)

The way I split these is into three groups of identities:

1. Addition/Subtraction Identities (A for Addition)

(A1: Addition of Sines)

sin(α±β)=sin(α)cos(β)+cos(α)sin(β)

(A2: Addition of Cosines)

cos(α+β)=cos(α)sin(β)sin(α)cos(β)

(A3: Addition of Tangents)

tan(α+β)=tan(α)+tan(β)1tan(α)tan(β),tan(αβ)=tan(α)tan(β)1tan(α)tan(β)

2. Multiplication/Division Identities (M for Multiplicaton)

(M1: Sine Double Angle)

sin(2θ)=2sin(θ)cos(θ)

(M2: Cosine Double Angle)

cos(2θ)=cos2(θ)sin2(θ)cos=12sin2(θ)cos=2cos2(θ)1

(M3: Tangent Double Angle)

tan(2θ)=2tan(θ)1tan(θ)

3. Squared Identities (S for Squared)

(Also known as the Pythagorean Identities)

(S1: Pythagorean Identity)

sin2(θ)+cos2(θ)=1

(S2: Pythagorean Tangent-Squared)

1+tan2(θ)=sec2(θ)

(S3: Pythagorean Cotangent-Squared)

cot2(θ)+1=csc2(θ)

Practice Problems

Problem 1: Summary from Today's Session

  1. From today: Verify the identity you have, that

    cos(2θ)=2cos2(θ)1

    Though we saw today how there are many, many ways you could verify this, using the identities you know, we also saw that probably the fastest way would be using the following steps (to go from the left-hand side

    cos(2θ) to the right-hand side
    2cos2(θ)1
    ):

    a. Break

    cos(2θ) into
    cos(θ+θ)
    , since this transforms the multiplication happening inside our
    cos
    into an addition, meaning that we can use our Addition Identities from above.

    cos(2θ)=cos(θ+θ)

    b. Use our identity (A2: Addition of Cosines) to turn this addition inside the cosine into an addition of trig functions without any additions happening inside of them. With

    θ as both our
    α
    and
    β
    values, we get

    cos(θ+θ)=cos(θ)cos(θ)sin(θ)sin(θ)=cos2(θ)sin2(θ).

    c. We're close, since we now have the

    cos2(θ) term that we wanted, but it came with an extra
    sin2(θ)
    term that we need to get rid of. So, since we see that the Pythagorean Identities are the ones that can help us when we have squared terms, let's take a look at the main one:

    cos2(θ)+sin2(θ)=1.

    d. We know we want to get rid of the

    sin2(θ) term in
    cos2(θ)sin2(θ)
    , so let's solve this Pythagorean identity for
    sin2(θ)
    , which will give us a way to rewrite this term:

    cos2(θ)+sin2(θ)=1sin2(θ)=1cos2(θ).

    So, we've derived a new identity, that anywhere we see

    sin2(θ) we can always rewrite it as
    1cos2(θ)
    ! So, let's rewrite the
    sin2(θ)
    term in the expression we had arrived at above:

    cos2(θ)sin2(θ)=cos2(θ)(1cos2(θ))=cos2(θ)1+cos2(θ)=2cos2(θ)1,

    and we're done! We started from

    cos(2θ), and used valid identities (and stuff derived from these identities) throughout to arrive at
    2cos2(θ)1
    , as desired.
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Problem 2: Lots of cancelling and simplifying

  1. The one you mentioned from class: Verify that

    sin(θ)1cos(θ)=cos(θ)cot(θ)cos(θ)cot(θ).

    In this case, as you mentioned in how you solved it, it isn't so bad identity-wise, instead it really requires a lot of algebra and cancelling stuff out. Starting from the right-hand side this time, we carry out the following steps (where I'm not including full explanations just because, you've already solved it! But, wanted to include so we have a bunch of examples in one place)

    a. Transform

    cot(θ) into
    cos(θ)sin(θ)
    :

    cos(θ)cot(θ)cos(θ)cot(θ)=cos(θ)cos(θ)sin(θ)cos(θ)cos(θ)sin(θ).

    b. Combine the

    cos(θ) terms in the denominator, so that it becomes a single fraction:

    cos(θ)cos(θ)sin(θ)cos(θ)cos(θ)sin(θ)=cos(θ)cos(θ)sin(θ)cos2(θ)sin(θ)

    c. Now we can "flip" the denominator, since division by a fraction is equivalent to multiplication by its reciprocal:

    cos(θ)cos(θ)sin(θ)cos2(θ)sin(θ)=(cos(θ)cos(θ)sin(θ))sin(θ)cos2(θ),

    d. And now we can multiply the big parentheses term through by

    sin(θ) to get

    cos(θ)sin(θ)cos(θ)cos2(θ),

    e. We can factor a

    cos(θ) out of the numerator to get

    cos(θ)(sin(θ)1)cos2(θ),

    f. And now the cosine in the numerator and one of the cosines in the denominator cancel out, giving us

    sin(θ)1cos(θ),

    as we wanted to show!

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Problem 3: Two Different Pathways to the Answer

  1. New problem! Verify the identity

    sec2(θ)1sec2(θ)=sin2(θ)

    Here, like I promised at the end of our session, we have an example where there are two totally-valid paths to the solution!

    Path 1:

    a. Use fraction rules to turn the left-hand side into a subtraction of two fractions:

    sec2(θ)1sec2(θ)=sec2(θ)sec2(θ)1sec2(θ)

    b. We see that the left fraction (before the minus sign) just turns into a

    1, while the right fraction (after the minus sign) is exactly our definition of
    cos2(θ)
    (since
    sec(θ)
    is defined as the reciprocal of
    cos(θ)
    ), giving us

    sec2(θ)sec2(θ)1sec2(θ)=1cos2(θ).

    Now, using the same Pythagorean identity reasoning we used in steps c and d of Problem 1 above, we know that we can rewrite

    1cos2(θ) as just
    sin2(θ)
    , giving us the result we wanted to verify!
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    Path 2:

    Here, at the beginning, we could use our (S3: Pythogrean Tangent Squared) identity from above, to turn that

    sec2(θ) into
    1+tan2(θ)
    :

    sec2(θ)1sec2(θ)=(1+tan2(θ))1sec2(θ).

    Note that we only apply this identity to the numerator, rather than both numerator and denominator, because the numerator is the part of the problem that has that pesky

    1 term we're trying to get rid of, to simplify the expression. Since we did that, we see that the
    1
    terms in the numerator indeed cancel, giving us

    (1+tan2(θ))1sec2(θ)=tan2(θ)sec2(θ).

    Now we can just use our definitions of these two functions to simplify our whole expression, write them just in terms of our two basic trig functions

    sin and
    cos
    :

    tan2(θ)sec2(θ)=sin2(θ)cos2(θ)1cos2(θ),

    and since dividing by a fraction is equivalent to multiplying by its reciprocal,

    sin2(θ)cos2(θ)1cos2(θ)=sin2(θ)cos2(θ)cos2(θ)1,

    and the

    cos2(θ) terms cancel, giving us just
    sin2(θ)
    , as we wanted to verify!
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Problem 4: Slightly harder problem requiring "conjugate multiplication"

  1. Verify

    cos(θ)1+sin(θ)=1sin(θ)cos(θ)

    a. Remember that "conjugate multiplication" is just that weird trick with fractions where you can sometimes get a simpler term by multiplying the numerator and the denominator by the "conjugate" of the denominator: here, let's look at the term on the left-hand side and see what happens if we multiply both the numerator and the denominator by the conjugate of the denominator, which is

    1sin(θ):

    cos(θ)1+sin(θ)=cos(θ)1+sin(θ)(1sin(θ)1sin(θ))=cos(θ)cos(θ)sin(θ)1sin2(θ)

    b. Even though it seemed like multiplying the denominator by its conjugate might give some really big gross messy term, in fact once we FOILed we found that a bunch of terms cancel, since

    (1+sin(θ))(1sin(θ))=1+sin(θ)sin(θ)sin2(θ),

    and the middle two terms cancel each other out.

    c. So, with the fraction we're left with, first note that by the same reasoning as in steps c and d of Problem 1 above, we can use our Pythagorean identity to rewrite the

    1sin2(θ) in the denominator as
    cos2(θ)
    :

    cos(θ)cos(θ)sin(θ)1sin2(θ)=cos(θ)cos(θ)sin(θ)cos2(θ).

    d. Now, we can also factor a

    cos(θ) term out of the numerator, to get

    cos(θ)cos(θ)sin(θ)cos2(θ)=cos(θ)(1sin(θ))cos2(θ).

    e. This means that the

    cos(θ) term in the numerator cancels out one of the
    cos(θ)
    terms in the denominator, so we're left with

    cos(θ)(1sin(θ))cos2(θ)=1sin(θ)cos2(θ),

    which is what we wanted to show!

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Additional Problems

Working these out on your own will be so so helpful, if you have additional time before the test!

  1. Verify

cot(θ)csc(θ)=cos(θ)

  1. Verify

sin2(θ)1tan(θ)sin(θ)tan(θ)=sin(θ)+1tan(θ)

  1. Verify

(1cos2(θ))(1+cot2(θ))=1