Sections 6.3 and 6.4: Adding and Subtracting Angles, Double-Angle Formulas
In this class, we’ll introduce a few new formulas for adding and subtracting two angles inside of a trig
function, like trig(α ± β). It’s easiest to do this for sin and cos.1
Caution: The HW has some problems with complements; look back at Section 5.3’s identities.
Cosine Addition/Subtraction:
Sine Addition/Subtraction:
sin(α ± β) = sin(α) cos(β) ± cos(α) sin(β)
cos(α ± β) = cos(α) cos(β) ∓ sin(α) sin(β)
• This formula has “mixed terms”: one sin and one
cos in each.
• The sides use the same operation.
• This formula “does not mix”: cosines come first,
then sines.
• The sides use opposite operations.
Why we use these: Say α and β are convenient angles in a problem or figure. These formulas help us find
values for the “combined angles”, i.e. trig(α + β) or trig(α − β).
• You’ll need four things for these identities: sine and cosine of both α and β! These values may come from
several places: points (x, y) on a terminal side, other trig values, reference + ASTC...
• Remember that all trigs are built from sine and cosine, e.g. tan(α ± β) = sin(α ± β)/ cos(α ± β).
Ex 1: Compute sin(π/12) and cos(π/12) exactly. (Use π/12 = π/3 − π/4.)
Hint: Think of α = π/3 and β = π/4 here. These are special angles!
Ex 2: Suppose cos(α) = 8/17 and tan(β) = 4/3, where α is in Quad IV and β is in Quad III. Find
(a) sin(α + β)
(b) cos(α + β)
(c) the quadrant of α + β
Starting out: You will need sin and cos for both α and β. Use reference triangles and ASTC! For example,
for α, your reference triangle has adj = 8 and hyp = 17, so you can find opp = 15.
Once you’ve done (a) and (b), use the signs to determine (c).
Ex 3: If sin(α) = −5/13 and tan(α) > 0, compute sin(α − π/6).
Hint: Think sin(α − β), where β = π/6. To find cos(α)’s sign, note sin(α) is negative and tan(α) is positive.
(What quadrant is α?)
Ex 4: Find the cosine of ∠P OQ in the figure.
Getting started: Let’s say θ = ∠P OQ. It’s not in
standard position (doesn’t start at positive x-axis.)
We draw two angles which are in standard position!
The bigger angle goes to Q; let’s call that angle α.
Similarly, say β goes to P . Therefore, θ = α − β, and
we want cos(α − β). Find sin and cos for α and β...
Double-Angle Formulas
When you use the addition formulas with β = α, you get formulas for sin(α + α) and cos(α + α).
Sine Double-Angle Formula: sin(2α) = 2 sin(α) cos(α)
Cosine Double-Angle Formula(s): cos(2α) = cos2 (α) − sin2 (α) .
• You can use sin2 + cos2 = 1 to convert these squares into other forms...
cos(2α) is: cos2 (α) − sin2 (α)
or
2 cos2 (α) − 1
or
1 − 2 sin2 (α)
The first formula is usually best, but sometimes you only want cosines or only want sines!
Ex 5: Suppose the angle T has (7, y) on its terminal side. Find cos(2T ) as a function of y.
Getting started: You have a point on T ’s terminal side (not 2T ’s!). Use it to find cos(T ) and sin(T ), which
are used in the double-angle formula(s). (Which of the three formulas do you find easiest to use? Does it
matter?)
1 If you want to know more about where these formulas come from, the textbook has some elaborate pictures for cos(α − β),
and then they determine the other identities through even/odd and complement identities. If you want more background, talk to
me outside of class.
Checking Identities with Angle Additions / Subtractions
Recall: If you want to know whether an identity is true, we usually simplify both sides by using identities or
basic algebra.
√
Ex 6: Is the formula sin(θ + π/4) = 2/2(sin(θ) + cos(θ)) correct?
Hint: Use sin(α + β) with α = θ and β = π/4.
Ex 7: Is the formula cos(u + v) − cos(u − v) = 2 sin(u) sin(v) correct?
Hint: Use identities for both cos(u + v) and cos(u − v). Simplify the result.
Solving Trig Equations With Multiple Unknown Angles
When a trig equation and multiple unknown angles are involved, our earlier tactics don’t quite suffice. We
have to turn our problem into something with ONE unknown angle!
We’ll see two main scenarios:
• Reversing an addition/subtraction identity: If you see multiplications with two angles, like trig(α)·trig(β),
this may apply. If you have two of these products, you may recognize it as the right side of one of the
identities from today!
Sample: sin(5t) cos(2t) − cos(5t) sin(2t) = 0. The left side comes from the sin(α − β) formula when you
have α = 5t and β = 2t, so the left side is just sin(5t − 2t) = sin(3t) = 0.
• Double-angle: If you have both 2x and x, use a double-angle identity on trig(2x). Hopefully, you get
something that either factors or uses the quadratic formula. Don’t forget there are three versions of
cos(2x), so think about which one works best in your equation!
Sample: cos(2x) − cos(x) = 0; we’ll rewrite cos(2x). Replacing it with cos2 (x) − sin2 (x) produces
unknowns of both cos(x) and sin(x), BUT if we replace with 2 cos2 (x) − 1, we get an equation with only
cosines! You get 2 cos2 (x) − 1 − cos(x) = 0, which is quadratic in cos(x).
Ex 8: Solve the equation cos(4t) cos(2t) = − sin(4t) sin(2t) for all t in [0, 2π).
Hint: Move everything to one side; you get either cos(α + β) or cos(α − β). (Which one?)
Ex 9: Solve the equation sin(u) − sin(2u) = 0 for u in [0, 2π).
Hint: After using double-angle, factor the result. You get two trig equations.
Ex 10: Solve the equation cos(2x) + cos(x) = 0 for x in [0, 2π).
Hint: The double-angle version with 2 cos2 (x) − 1 is most useful, so you get a quadratic in cos.
More Practice (outside of class)
These kinds of problems show up in HW and WebQuizzes. Answers can be found in a handout on my
website.
Ex 11: Express the following as a trig function of a single angle: sin(63◦ ) cos(19◦ ) + cos(63◦ ) sin(19◦ ).
Hint: You can tell this came from sin(α ± β) formula because of the mix of sin and cos parts in each term,
but how do you figure out whether it uses + or −?
Ex 12: If cot(θ) = 4/3 and 180◦ < θ < 270◦ , find the values of sin(θ), cos(θ), and sin(2θ).
Getting started: You have cot(θ) and θ’s quadrant. This means you should be able to get sin(θ) and cos(θ)
(draw a reference triangle), which are used in the double-angle formula for sin(2θ).
Factoring by grouping: You may get a problem that requires this factoring technique. If you have four terms,
you put them in groups of two, and try to factor from each group. If you’re lucky, both groups have a common
factor!
Sample: We break down x3 − 2x2 + 7x − 14 as
x3 − 2x2 + 7x − 14 = (x3 − 2x2 ) + (7x − 14) = x2 (x − 2) + 7(x − 2)
Since (x − 2) is a common factor to both, we can factor it out and get (x − 2)(x2 + 7).
Ex 13: Solve sec(x) tan(x) + tan(x) − sec(x) − 1 = 0 for all solutions x in [0, 2π).
Hint: The first group of two has a sec(x) factor, and the second group has a −1 factor. You should be able
to rewrite this as
(tan(x) − 1)(sec(x) + 1) = 0
This gives you two basic trig equations: tan(x) = 1 and sec(x) = −1.