Download Andrew`s handout on trig identities

Andrew’s handout
1
1.1
Trig identities
Fundamental identities
These are the most fundamental identities, in the sense that you should probably memorize these and use them to
derive the rest (or, if you prefer, memorize all the identities, including these). There’s not really a good way to derive
these identities; all the derivations are (in my opinion) much harder to remember than the identities themselves are.
Identity 1 (The fundamental trig identity, a.k.a. the Pythagorean identity).
sin2 (x) + cos2 (x) = 1 for any x ∈ R
Identity 2 (Angle addition formulas).
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
cos(x + y) = cos(x) cos(y) − sin(x) sin(y)
1.2
Other identities coming from the Pythagorean identity
These are really simple, but I’m repeating them for good measure.
Take the Pythagorean identity, and divide both sides by cos2 (x):
sin2 (x) + cos2 (x) = 1
sin2 (x) cos2 (x)
1
+
=
cos2 (x) cos2 (x) cos2 (x)
tan2 (x) + 1 = sec2 (x)
Or, solve for sin or cos (sometimes useful to solve a trig integral by u-substitution):
q
q
sin(x) = ± 1 − cos2 (x)
cos(x) = ± 1 − sin2 (x)
1.3
Consequences of the angle addition formulas
First, the double angle formulas:
sin(2x) = sin(x + x) = 2 sin(x) cos(x)
cos(2x) = cos(x + x) = cos2 (x) − sin2 (x)
The double angle formula for cos can be written in a few different ways:
cos(2x) = cos2 (x) − sin2 (x)
= cos2 (x) − (1 − cos2 (x))
= 2 cos2 (x) − 1
Or,
cos(2x) = cos2 (x) − sin2 (x)
= (1 − sin2 (x)) − sin2 (x)
= 1 − 2 sin2 (x)
1
These two forms give us a nice way of rewriting powers of sin or cos (just solve for cos2 (x) and sin2 (x) in the above
identities):
1 + cos(2x)
2
1
−
cos(2x)
sin2 (x) =
2
cos2 (x) =
1.4
Translations
The identities in this section can all be derived graphically, by shifting the graphs of sin and cos, or by reflecting across
the y-axis, or by similar means.
For example, if you graph sin and cos on the same graph, it should be clear that the curves are identical except that sin
is shifted right by π2 units:
y
sin(x)
cos(x)
1
0.5
x
−2π
−π
π
2π
−0.5
−1
What this means, is that any time we want to calculate sin(x), we can instead calculate cos(x − π2 ):
y
sin(x)
cos(x)
1
0.5
x
−2π
−π
π
2π
−0.5
−1
In other words, sin(x) = cos(x − π2 ). We can check this using the angle addition formula for cos(x):
cos(x −
π
π
π
) = cos(x) cos(− ) − sin(x) sin(− ) = cos(x) · 0 − sin(x) · (−1) = sin(x)
2
2
2
2
Another important pair of identities captures the fact that π is half the wavelength of sin and cos, so sin(x + π) =
− sin(x) and cos(x + π) = − cos(x). Again, you can check these identities with the angle addition formulas, but I find
it’s easiest to just remember these as “shifting by half the wavelength negates the sine wave”.
y
1
sin(x)
sin(x + π)
0.5
x
−2π
−π
π
2π
−0.5
−1
Finally, looking at the graphs of sin and cos, you can see that sin(x) is an odd function and cos(x) is an even function.
In particular, sin(−x) = − sin(x):
y
sin(x)
sin(−x)
1
0.5
x
−2π
−π
π
2π
−0.5
−1
and cos(−x) = cos(x):
y
cos(x)
cos(−x)
1
0.5
x
−2π
−π
π
2π
−0.5
−1
3
2
2.1
Simplifying expressions like sin(arctan(x))
Geometric approach
This is the easiest way to simplify these types of expressions for trig functions. Note though that this method doesn’t
work (or at least not easily) for hyperbolic functions.
We’ll use the geometric definitions of sin and friends, as captured by the mnemonic “SOH-CAH-TOA”. We’ll draw a
right triangle so that one of its acute angles represents the inner inverse trig function, then solve the triangle to compute
the value of the outer trig function. If that sounds confusing, don’t worry, it’s much easier to follow along with an
example.
Example 1. Simplify: sin(arctan(x)).
Proof. For convenience, write θ = arctan(x), so that tan θ = x = 1x . The reason for the silly notation
the definition of tan as the ratio of the lengths of the opposite and adjacent sides in a right triangle:
x
x
1
is to connect to
θ
1
Note that we could have assigned the side lengths to be anything, as long as their ratio was equal to tan(θ ) = x. So,
for example, we could have drawn this triangle instead:
2x
θ
2
Of course, there’s no good reason for doing that, so we’ll keep the original triangle.
Now, we can easily solve for the length of the third side:
√
1 + x2
x
θ
1
And now, using the ratio definition of sin, we can read off
x
sin(θ ) = √
1 + x2
Since we defined θ = arctan(x), we have shown that
x
sin(arctan(x)) = √
1 + x2
2.2
Algebraic approach
This approach also works with hyperbolic functions. The idea is to use trig identities to replace the inner trig function
by the outer (inverse) trig function.
Example 2. Simplify: sin(arctan(x)).
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Solution. We want to replace the outer sin by tan, to make use of the fact that tan(arctan(x)) = x.
We know the following:
sin2 (x) + cos2 (x) = 1
Since we want to get a relation between tan(x) and sin(x), we divide by sin2 (x):
1+
1
1
=
tan2 (x) sin2 (x)
(Often, we write the above identity with cot and csc instead of
involves sin and tan, it’s better to use sin and tan explicitly.)
1
tan
and
1
sin ,
but since the expression we’re simplifying
We want to use this trig identity to replace sin(arctan(x)) by something involving tan(arctan(x)). From the identity,
sin2 (x) = 1 +
1
tan2 (x)
−1
so
sin2 (arctan(x)) =
and finally
2.3
=
tan2 (x) + 1
tan2 (x)
−1
=
tan2 (x)
tan2 (x) + 1
tan2 (arctan(x))
x2
=
tan2 (arctan(x)) + 1 x2 + 1
x
sin(arctan(x)) = √
2
x +1
Extension to hyperbolic functions
The advantage of the algebraic method from section 2.2 is that in extends easily to hyperbolic functions (the geometric
approach doesn’t extend so well, since the geometry behind hyperbolic functions is tied to hyperbolas instead of
more-familiar circles).
Example 3. Simplify: sinh(artanh(x)) (where artanh(x) denotes the inverse hyperbolic tangent)
Solution. The basic hyperbolic relation is
cosh2 (x) − sinh2 (x) = 1
Divide both sides by sinh2 (x), by analogy to the example from section 2.2:
1
1
−1 =
2
tanh (x)
sinh2 (x)
Thus,
sinh2 (x) =
−1 −1
1
1 − tanh2 (x)
tanh2 (x)
−
1
=
=
2
2
tanh (x)
tanh (x)
1 − tanh2 (x)
So, replacing x by artanh(x), we have:
sinh2 (artanh(x)) =
Thus,
tanh2 (artanh(x))
x2
=
1 − tanh2 (artanh(x)) 1 − x2
x
sinh(artanh(x)) = √
1 − x2
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3
3.1
Trig and inverse trig derivatives and integrals
Trig Derivatives
I’m going to assume you know that dxd [sin(x)] = cos(x) and dxd [cos(x)] = − sin(x). Then, it’s easy to compute the rest
of the trig derivatives if you don’t have them memorized. For example,
d sin(x)
1
d
cos(x)(cos(x)) − sin(x)(− sin(x))
[tan(x)] =
=
= sec2 (x)
=
dx
dx cos(x)
cos2 (x)
cos2 (x)
d
1
1
d
=− 2
[csc(x)] =
cos(x) = − cot(x) csc(x)
dx
dx sin(x)
sin (x)
3.2
Inverse trig derivatives
These are slightly harder than the trig derivatives, but only slightly. Basically, use the definition of the inverse trig
function and implicit differentiation:
Example 4. Find dxd [arccos(x)].
Solution. Let y = arccos(x), so that cos(y) = x. We want to find
dy
dx .
Let’s use implicit differentiation:
d
d
[cos(y)] =
[x]
dx
dx
dy
− sin(y) = 1
dx
dy
1
=−
dx
sin(y)
dy
1
=−
dx
sin(arccos(x))
The only thing left to do is to find the relation
sin(arccos(x)) =
p
1 − x2
d
1
[arccos(x)] = − √
dx
1 − x2
=⇒
See section 2 for details.
3.3
Trig integrals
R
R
Two (well, really four) trig integrals deserve
special mention: tan(x) dxR = − ln(cos(x))+C and sec(x) dx = ln(tan(x)+
R
sec(x)) +C (and the similar integrals cot(x) dx = ln(sin(x)) +C and csc(x) dx = − ln(cot(x) + csc(x)) +C).
The tangent/cotangent integrals are easy:
Z
sin(x)
dx
cos(x)
Z
− du
=
u
= − ln(u) +C
Z
tan(x) dx =
u = cos(x), du = − sin(x) dx
= − ln(cos(x)) +C
(if you prefer ln(sec(x)) to − ln(cos(x)))
= ln(sec(x)) +C
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R
R
The other non-obvious trig integral is sec(x) dx (or csc(x) dx). You might just want to memorize
Z
sec(x) dx = ln(tan(x) + sec(x)) +C
On the other hand, in case you’re interested, here’s a reasonably-memorable calculation (courtesy of Wikipedia,
https://en.wikipedia.org/wiki/Integral_of_the_secant_function):
Z
cos(x)
dx
cos2 (x)
Z
cos(x)
=
dx
1 − sin2 (x)
Z
dt
=
1 − t2
Z 1
1
1
=
+
dt
2
1+t 1−t
1
1
= ln(1 + t) − ln(1 − t) +C
2 2
1+t
1
+C
= ln
2
1−t
1
1 + sin(x)
= ln
+C
2
1 − sin(x)
1
(1 + sin(x) 1 + sin(x)
= ln
·
+C
2
1 − sin(x) 1 + sin(x)
s
(1 + sin(x))2
= ln
1 − sin2 (x)
s
(1 + sin(x))2
= ln
cos2 (x)
1 + sin(x)
= ln
cos(x)
Z
sec(x) dx =
t = sin(x), dt = cos(x) dx
By partial fractions; easy.
Note: already a reasonable answer.
Multiply by the conjugate – often useful!
= ln(sec(x) + tan(x))
R
Before you give up and just memorize, sec(x) dx = ln(sec(x) + tan(x)), you might want to try working through this
calculation by hand (without just copying my notes or the Wikipedia page!) to see if that’s easier for you to remember.
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(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
(19)
(20)
Identity
sin2 (x) + cos2 (x) = 1
sin2 (x) = 1 − cos2 (x)
cos2 (x) = 1 − sin2 (x)
tan2 (x) + 1 = sec2 (x)
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
cos(x + y) = cos(x) cos(y) − sin(x) sin(y)
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2 (x) − sin2 (x)
cos(2x) = 2 cos2 (x) − 1
cos(2x) = 1 − 2 sin2 (x)
cos2 (x) = 12 (1 + cos(2x))
sin2 (x) = 12 (1 − cos(2x))
sin(x − π2 ) = cos(x)
sin(x + π) = − sin(x)
cos(x + π) = − cos(x)
cos(x − π2 ) = − sin(x)
cos(x + π2 ) = sin(x)
sin(x + π2 ) = − cos(x)
tan(x ± π2 ) = − tan(x)
tan(x + π) = tan(x)
Mnemonic
Memorize
Restatement of (1)
Restatement of (1)
Divide both sides of (1) by cos2 (x)
Memorize
Memorize
Derive (easily!) from (5)
Derive (easily!) from (6)
Combine (8) and (2)
Combine (8) and (3)
Solve (9) for cos2 (x)
Solve (10) for sin2 (x)
Look at the graphs (see above)
Half-wavelength
Half-wavelength
Combine (13) and (14) (try it!)
Restatement of (13)
Restatement of (16)
Combine (17)+(18) or (13)+(16)
Combine (14) and (15)
Table 1: Trig identities and mnemonics
(1)
(2)
(3)
(4)
Result
derivative of tan(x), sec(x), etc.
derivative of arctan(x), etc.
R
tan(x) dx = ln(sec(x)) +C
R
sec(x) dx = ln(sec(x) + tan(x)) +C
Method
write as a fraction in sin(x) and cos(x); use quotient rule
write e.g. tan(y) = x; find dy/ dx via implicit differentiation
write as a fraction sin(x)/ cos(x); sustitute u = cos(x)
Memorize, or use sec(x) = cos(x)/(1 − sin2 (x)),
substitute u = sin(x), and then use partial fractions
Table 2: Calculus with trigonometric functions
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