Andrew’s handout

1 Trig identities

1.1 Fundamental identities

These are the most fundamental identities, in the sense that you should probably memorize these and use them toderive the rest (or, if you prefer, memorize all the identities, including these). There’s not really a good way to derivethese 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)

+cos2(x)cos2(x)

=1

cos2(x)

tan2(x)+1 = sec2(x)

Or, solve for sin or cos (sometimes useful to solve a trig integral by u-substitution):

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) = 2sin(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))

= 2cos2(x)−1

Or,

cos(2x) = cos2(x)− sin2(x)

= (1− sin2(x))− sin2(x)

= 1−2sin2(x)

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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 aboveidentities):

cos2(x) =1+ cos(2x)

2

sin2(x) =1− cos(2x)

2

1.4 Translations

The identities in this section can all be derived graphically, by shifting the graphs of sin and cos, or by reflecting acrossthe 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 sinis shifted right by π

2 units:

−2π −π π 2π

−1

−0.5

0.5

1

x

ysin(x)cos(x)

What this means, is that any time we want to calculate sin(x), we can instead calculate cos(x− π

2 ):

−2π −π π 2π

−1

−0.5

0.5

1

x

ysin(x)cos(x)

In other words, sin(x) = cos(x− π

2 ). We can check this using the angle addition formula for cos(x):

cos(x− π

2) = cos(x)cos(−π

2)− sin(x)sin(−π

2) = cos(x) ·0− sin(x) · (−1) = sin(x)

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 findit’s easiest to just remember these as “shifting by half the wavelength negates the sine wave”.

−2π −π π 2π

−1

−0.5

0.5

1

x

ysin(x)

sin(x+π)

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):

−2π −π π 2π

−1

−0.5

0.5

1

x

ysin(x)

sin(−x)

and cos(−x) = cos(x):

−2π −π π 2π

−1

−0.5

0.5

1

x

ycos(x)

cos(−x)

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2 Simplifying expressions like sin(arctan(x))

2.1 Geometric approach

This is the easiest way to simplify these types of expressions for trig functions. Note though that this method doesn’twork (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 aright triangle so that one of its acute angles represents the inner inverse trig function, then solve the triangle to computethe value of the outer trig function. If that sounds confusing, don’t worry, it’s much easier to follow along with anexample.Example 1. Simplify: sin(arctan(x)).

Proof. For convenience, write θ = arctan(x), so that tanθ = x = x1 . The reason for the silly notation x

1 is to connect tothe definition of tan as the ratio of the lengths of the opposite and adjacent sides in a right triangle:

θx

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:

θ

2x

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:

θx

√1+ x2

1

And now, using the ratio definition of sin, we can read off

sin(θ) =x√

1+ x2

Since we defined θ = arctan(x), we have shown that

sin(arctan(x)) =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 functionby 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

tan2(x)=

1sin2(x)

(Often, we write the above identity with cot and csc instead of 1tan and 1

sin , but since the expression we’re simplifyinginvolves sin and tan, it’s better to use sin and tan explicitly.)

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

=

(tan2(x)+1

tan2(x)

)−1

=tan2(x)

tan2(x)+1

so

sin2(arctan(x)) =tan2(arctan(x))

tan2(arctan(x))+1=

x2

x2 +1

and finallysin(arctan(x)) =

x√x2 +1

2.3 Extension to hyperbolic functions

The advantage of the algebraic method from section 2.2 is that in extends easily to hyperbolic functions (the geometricapproach doesn’t extend so well, since the geometry behind hyperbolic functions is tied to hyperbolas instead ofmore-familiar circles).

Example 3. Simplify: sinh(artanh(x)) (where artanh(x) denotes the inverse hyperbolic tangent)

Solution. The basic hyperbolic relation iscosh2(x)− sinh2(x) = 1

Divide both sides by sinh2(x), by analogy to the example from section 2.2:

1tanh2(x)

−1 =1

sinh2(x)

Thus,

sinh2(x) =(

1tanh2(x)

−1)−1

=

(1− tanh2(x)

tanh2(x)

)−1

=tanh2(x)

1− tanh2(x)

So, replacing x by artanh(x), we have:

sinh2(artanh(x)) =tanh2(artanh(x))

1− tanh2(artanh(x))=

x2

1− x2

Thus,sinh(artanh(x)) =

x√1− x2

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3 Trig and inverse trig derivatives and integrals

3.1 Trig Derivatives

I’m going to assume you know that ddx [sin(x)] = cos(x) and d

dx [cos(x)] =−sin(x). Then, it’s easy to compute the restof the trig derivatives if you don’t have them memorized. For example,

ddx

[tan(x)] =d

dx

[sin(x)cos(x)

]=

cos(x)(cos(x))− sin(x)(−sin(x))cos2(x)

=1

cos2(x)= sec2(x)

ddx

[csc(x)] =d

dx

[1

sin(x)

]=− 1

sin2(x)cos(x) =−cot(x)csc(x)

3.2 Inverse trig derivatives

These are slightly harder than the trig derivatives, but only slightly. Basically, use the definition of the inverse trigfunction and implicit differentiation:Example 4. Find d

dx [arccos(x)].

Solution. Let y = arccos(x), so that cos(y) = x. We want to find dydx . Let’s use implicit differentiation:

ddx

[cos(y)] =d

dx[x]

−sin(y)dydx

= 1

dydx

=− 1sin(y)

dydx

=− 1sin(arccos(x))

The only thing left to do is to find the relation

sin(arccos(x)) =√

1− x2 =⇒ ddx

[arccos(x)] =− 1√1− x2

See section 2 for details.

3.3 Trig integrals

Two (well, really four) trig integrals deserve special mention:∫

tan(x)dx=− ln(cos(x))+C and∫

sec(x)dx= ln(tan(x)+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:∫tan(x)dx =

∫ sin(x)cos(x)

dx

=∫ −du

uu = cos(x),du =−sin(x)dx

=− ln(u)+C

=− ln(cos(x))+C

= ln(sec(x))+C (if you prefer ln(sec(x)) to − ln(cos(x)))

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The other non-obvious trig integral is∫

sec(x)dx (or∫

csc(x)dx). You might just want to memorize∫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):∫

sec(x)dx =∫ cos(x)

cos2(x)dx

=∫ cos(x)

1− sin2(x)dx

=∫ dt

1− t2 t = sin(x),dt = cos(x)dx

=12

∫ ( 11+ t

+1

1− t

)dt By partial fractions; easy.

=12

ln(1+ t)− 12

ln(1− t)+C

=12

ln(

1+ t1− t

)+C

=12

ln(

1+ sin(x)1− sin(x)

)+C Note: already a reasonable answer.

=12

ln((1+ sin(x)1− sin(x)

· 1+ sin(x)1+ sin(x)

)+C Multiply by the conjugate – often useful!

= ln

√(1+ sin(x))2

1− sin2(x)

= ln

√(1+ sin(x))2

cos2(x)

= ln(

1+ sin(x)cos(x)

)= ln(sec(x)+ tan(x))

Before you give up and just memorize,∫

sec(x)dx = ln(sec(x)+ tan(x)), you might want to try working through thiscalculation 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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Identity Mnemonic(1) sin2(x)+ cos2(x) = 1 Memorize(2) sin2(x) = 1− cos2(x) Restatement of (1)(3) cos2(x) = 1− sin2(x) Restatement of (1)(4) tan2(x)+1 = sec2(x) Divide both sides of (1) by cos2(x)(5) sin(x+ y) = sin(x)cos(y)+ cos(x)sin(y) Memorize(6) cos(x+ y) = cos(x)cos(y)− sin(x)sin(y) Memorize(7) sin(2x) = 2sin(x)cos(x) Derive (easily!) from (5)(8) cos(2x) = cos2(x)− sin2(x) Derive (easily!) from (6)(9) cos(2x) = 2cos2(x)−1 Combine (8) and (2)

(10) cos(2x) = 1−2sin2(x) Combine (8) and (3)(11) cos2(x) = 1

2 (1+ cos(2x)) Solve (9) for cos2(x)(12) sin2(x) = 1

2 (1− cos(2x)) Solve (10) for sin2(x)(13) sin(x− π

2 ) = cos(x) Look at the graphs (see above)(14) sin(x+π) =−sin(x) Half-wavelength(15) cos(x+π) =−cos(x) Half-wavelength(16) cos(x− π

2 ) =−sin(x) Combine (13) and (14) (try it!)(17) cos(x+ π

2 ) = sin(x) Restatement of (13)(18) sin(x+ π

2 ) =−cos(x) Restatement of (16)(19) tan(x± π

2 ) =− tan(x) Combine (17)+(18) or (13)+(16)(20) tan(x+π) = tan(x) Combine (14) and (15)

Table 1: Trig identities and mnemonics

Result Method(1) derivative of tan(x), sec(x), etc. write as a fraction in sin(x) and cos(x); use quotient rule(2) derivative of arctan(x), etc. write e.g. tan(y) = x; find dy/dx via implicit differentiation(3)

∫tan(x)dx = ln(sec(x))+C write as a fraction sin(x)/cos(x); sustitute u = cos(x)

(4)∫

sec(x)dx = ln(sec(x)+ tan(x))+C 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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