14.2 - Trigonometric identities

Identities are statements that are always true, as opposed to equations which are only true under certain conditions. For example 3x + 2x = 5x is an identity which is always true whereas 3x = 15 is an equation (or more precisely, a conditional equation) which is only true if x = 5.

Trigonometric identities are identities that contains trigonometric functions such as sin, cos or tan. Each identity describes some property of some trigonometric function. These identities can be used to:

Simplify trigonometric expressions.
Prove that two trigonometric expressions are equivalent. This may lead to new insights into some application.
Help solve trigonometric equations that arise in various applications.

We will state about twenty trigonometric identites. You can refer to books such as the “Handbooks of Mathematical Functions”, by Abramowitz and Stegun for many more. To manage them we will organize them into groups and discuss each group as we go.

The basic identities

1a.

1b.

These identities show how the sin, cos and tan functions are related.

Identity (1a) follows from the definitions of sin, cos and tan. To derive it simply divide the ratio for sin by the ratio for cos and the result is the ratio for tan:

Identity (1b) states that a cos wave is just a sin wave shifted to the left by a phase angle of π/2 or 90°.

The negative angle identities

2a.

sin(−α) = −sin(α)

2b.

cos(−α) = cos(α)

2c. tan(−α) = −tan(α)

These identities describe the left-right symmetry of the sin, cos and tan curves. For example, identity (2a) says that the height of the sin curve for a negative angle is the negative of the height of the sin curve for the corresponding positive angle. (See the picture below.) These identities are usually used to get rid of negative angles in favor of positive angles.

Identities for the sum and difference of two angles

3a.

sin(α ± β) = sin(α)·cos(β) ± cos(α)·sin(β)

3b.

3c.

These identities are usually used to convert a trigonometric function of a sum or difference of angles α and β into the trigonometric functions of the separate angles α and β.

These are actually 6 identities, 3 come from using the upper signs and 3 come from using the lower signs. For example identity (3c) with the lower signs reads:

A common mistake is to believe that, for example:

sin(α + β) = sin(α)+ sin(β)

This is not correct! It would only be correct if the sin curve was a straight line through the origin, instead of a wavy curve. The previous identities show the correct way to break up the sum of the angles.

In the last step we substituted in trigonometric functions of the two small right triangles.

Identity (3a) with the upper sign, namely:

sin(α + β) = sin(α)·cos(β) + cos(α)·sin(β)

which we just proved, is the most important identity of this group. Any other identity in this group (and in fact almost any trigonometric identity) can be derived by modifying it to look like this one. For example:

The identity for sin(α − β) can be derived by writing it as sin(α + (−β)), applying identity (3a) and then using the negative angle identities to clean up.
The identity for cos(α + β) can be derived by using identity (1b) to write it as a sin with a phase shift:
sin((α + 90°) + β),
then applying identity (3a) and then using (1b) again to clean up.
The identity for tan(α + β) can be derived by dividing identity (3a) by (3b).

Pythagoras’ theorem

Recall that identity (3b) reads

. If we use the lower sign and let β equal α then we get this:

4. sin²(α) + cos²(α) = 1

This is Pythagoras’ theorem in the form of a trigonometric identity. Note that sin²(α) is the standard shorthand for (sin(α))², and cos²(α) is the standard shorthand for (cos(α))². The reason for not placing the exponent at the end is to make it clear that it is not α that is squared, rather it is the sin of α that is squared.

Besides using identity (3b) there are many other ways to derive identity (4). Another way is to simply construct a right triangle with hypotenuse 1 and note that the base and altitude are given by the cosine and sine of θ and then write down the original form of Pythagoras’ theorem for this triangle.

The double angle identities

Identities (3a), (3b) and (3c) read:

sin(α ± β) = sin(α)·cos(β) ± cos(α)·sin(β)

If we use the upper signs in these identities and let β equal α we get:

5a.

sin(2α) = 2·sin(α)·cos(α)

5b.

cos(2α) = cos²(α) − sin²(α)
= 1 − 2 sin²(α)
= 2 cos²(α) − 1

5c.

The second and third forms of identity (5b) result from using Pythagoras’ theorem on the first form. They are the preferred forms because they only involve the sin or the cos, not both.

These identities are used to convert a trigonometric function of twice an angle into a trigonometric function of the angle itself.

The half angle identities

If we solve the second form of identity (5b) for sin²(α) and the third form for cos²(α) then we get these two identities:

If we now change the name of the angle α to be A/2 then these become the so-called half angle identities:

6a.

6b.

These identities are used to convert a trigonometric function of half an angle into a trigonometric function of the angle itself.

The product identities

If we add or subtract identities (3a) and (3b) in various combinations then we get the so-called product identites:

7a.

sin(α)·sin(β) = ½[cos(α − β) − cos(α + β)]

7b.

sin(α)·cos(β) = ½[sin(α + β) + sin(α − β)]

7c. cos(α)·cos(β) = ½[cos(α + β) + cos(α − β)]

These are useful to change products of trigonometric functions into sums of trigonometric functions or vice versa.

Applications

One application of trigonometric identities is to prove that two trigonometric expressions are equivalent. This may lead to new insights. In carrying out the proof you must not move expressions from one side to the other. Instead you use trigonometric identities to modify the left side or the right side or both sides until they look the same.

Strategies:

If sums, differences or multiples of angles occur on one side and not the other, then use an identity to separate the angles.
If sins and cosines occur on one side and not the other, then use an identity to convert the other side to sines and cosines.

Notes:

1) Replace sec(x) by

and csc(x) by

. Since the angles on the RHS (right-hand-side) are 4x and x, replace 5x by 4x + x.

2) Use identity (3a) on the numerator.

3) Break into 2 simpler fractions.

Notes:

1) Since the angle on the LHS is α, replace angle 2α on the RHS using identity (5a).

2) The RHS has a denominator cos(α) + sin(α). The easiest way to get the same denominator on the LHS is to multiply and divide the LHS by cos(α) + sin(α).

3) Use Pythagoras’ theorem.

Example 3. Addition of sine and cosine waveforms: (a) Prove the identity

, where on the right-hand-side C and φ are given by

(b) Use the result to find the sum of the two sinusoidal waveforms

y₁ = 3 sin(θ) and y₂ = 4 cos(θ).

Solution part (a): We will work on the right-hand-side of this identity and make it look like the left-hand-side. The equation for C is Pythagoras’ theorem and this means that we can draw a right triangle to describe the relationship between A, B and C. Furthermore, the equation for φ means that B must be the altitude of this triangle and A must be the base. The triangle is shown to the right. Note for future reference that this triangle also shows that

A = C cos(φ) and B = C sin(φ).

Notes:

1) Use identity (3a) to break apart angles θ and φ.

2) Distribute and change the order of the factors in the two resulting terms.

3) Use the fact (mentioned above) that A = C cos(φ) and B = C sin(φ).

Solution part (b): The sum of the two waveforms is y = 3 sin(θ) + 4 cos(θ). In the identity proven in part (a), let A = 3 and B = 4. This gives

so the sum of the waveforms is

y =

The graph to the right shows the separate waveforms and the waveform that is their sum.

Example 4. Beats:

(a) Prove the identity:

(b) Define the two sinusoidal waves:

Here are their graphs:

Notice that they have slightly different angular velocities (19 and 21). One result of this is that at certain times, for example at time t = 0, the waves are in phase (that is, going the same way), and at other times, for example at time t = 1.4, the waves are 180° out of phase, (that is, one is going positive when the other is going negative). Use the identity from part (a) to show that when the two sinusoidal waves are added, their sum, y₁ + y₂, produces this graph:

Solution part (a):

Notes:

1) The RHS has angles t and 20 t. To get the same angles on the LHS write 21 t as 20 t + 1 t and 19 t as 20 t − 1 t.

2) Apply identity (7b). In this case let α = 20 t and β = 1 t.

Solution part (b): Thus the sum of the waves can also be written as a product:

y₁ + y₂ = 1.0 cos(t) · sin(20 t).

The factor 1.0 cos(t) in this product plays the role of a slowly varying amplitude for the rapidly varying oscillation sin(20 t). (In other words imagine a sinusoidal wave sin(20 t), but that at certain times it has a small amplitude and at other times it has a large amplitude.) The periodic constructive and destructive interference is known as the beat phenomenon in acoustics.

Example 5. Electric power: When an electric current flows through a resistor (such as a toaster or light bulb) it produces heat or light or some other form of power. The power produced is given by the formula:

P = i² R,

where P is the power produced measured in watts, i is the electric current flowing through the resistor in amperes and R is the resistance of the resistor in ohms.

There are two types of electric current in common use, namely direct current and alternating current:

Direct current (DC) is current that is constant over time, such as that produced by a battery. For this example we will use a direct current of 0.7071 amperes:
i = 0.7071 amps
Alternating current (AC) is current that flows back and forth through the resistor and is described by a sinusoidal function of time. For this example we will use an alternating current with an amplitude of 1.0 amperes:
i = 1.0 sin(t) amps

The graph on the left shows the direct current and the graph on the right shows the alternating current:

Problem: Show that this direct current of 0.7071 amps and this alternating current with amplitude 1.0 amps both produce the same amount of power in a 1000Ω resistor, namely 500W, when averaged over time.

Solution: The basic idea is shown in these graphs where we have plotted the quantity i² corresponding to the above graphs of i:

On the left we see that squaring a constant current i = 0.7071 gives a constant i² = 0.5. When this i² is multiplied by R = 1000Ω then this gives a constant power P = 500 Watts.

On the right we see that squaring a sinusoidal current i = 1.0 sin(t) gives another wave shape i² = 1.0 sin²(t), with these features:

Both when i = + 1 and when i = −1 then i² = 1, and whenever i = 0 then i² = 0 also.
The wave shape looks like a cosine curve turned upside-down and raised to an average height of 0.5.

When this wave shape i² is multiplied by R = 1000Ω then this gives exactly the same wave shape for the power. When the average height of i², namely 0.5, is multiplied by R = 1000Ω then this gives an average power P = 500 Watts, as required.

Here is the proof that the average height of the i² curve is ½.

Start with: i = 1.0 sin(t) and square it: i² = 1.0 sin²(t).
Use identity (6a), modified to read .
Interpretation: .
The last term causes i² to have an average value of ½.

Algebra Coach Exercises