14.2 - Trigonometric identities

We begin this section by stating about 20 basic trigonometric identites. You can refer to books such as the “Handbook of Mathematical Functions”, by Abramowitz and Stegun for many more. To understand them we will organize them into 9 groups and discuss each group.

Then we finish this section with 7 examples.

Identities and equations compared: An identity is a statement that is always true, whereas an equation is only true under certain conditions. For example

3x + 2x = 5x

is an identity that is always true, no matter what the value of x, whereas

3x = 15

is an equation (or more precisely, a conditional equation) that is only true if x = 5.

A Trigonometric identity is an identity that contains the trigonometric functions sin, cos, tan, cot, sec or csc. Trigonometric identities can be used to:

Simplify trigonometric expressions.
Solve trigonometric equations.
Prove that one trigonometric expression is equivalent to another, so that we can replace the first expression by the second expression. The second expression can give us new insights into some application that the first one doesn’t show.

The reciprocal identities

1a.

1b.

1c.

This group of identities states that csc and sin are reciprocals, that sec and cos are reciprocals, and that cot and tan are reciprocals. They follow directly from the definitions of the trigonometric functions. (Click here to see them again.) Many mathematicians consider them to be merely basic simplifications to be used to get rid of csc, sec and cot in favor of sin, cos and tan.

The negative angle identities

2a.

cos (−θ ) = cos ( θ )

2b.

sin (−θ ) = − sin ( θ )

2c. tan (−θ ) = − tan ( θ )

These identities describe the left-right symmetry of the cos, sin and tan curves. Many mathematicians consider these identities to be just basic simplifications to be used to get rid of negative angles inside a cos, sin or tan.

Graphically, identity (2a) says that the height of the cos curve for a negative angle is the same as the height of the cos curve for the corresponding positive angle. Any curve having this property is said to have even symmetry.

Identity (2b) says that the height of the sin curve for a negative angle is the negative of the height for the corresponding positive angle. Any curve having this property is said to have odd symmetry.

Identity (2c) says that the tan curve also has odd symmetry.

The “left shift by 90° ” identities

3a.

3b.

3c.

The graph on the left explains identity (3a). It shows that shifting a sin curve to the left by 90° (or by π/2 radians or ¼ cycle) produces a cos curve. The graph on the right explains identity (3b). It shows that shifting a cos curve to the left by 90° (or by π/2 radians or ¼ cycle) produces an upside-down sin curve.

Identity (3c) can be proven by using identities (3a) and (3b) and a bit of algebra.

These identities are also considered to be basic simplifications and are used to get rid of the 90° shift. By using them repeatedly they can be used to get rid of shifts of 180°, 270°, 360°, etc.

Pythagoras’ theorem

4. sin²(θ) + cos²(θ) = 1

This is the trigonometric form of Pythagoras’ theorem.

Note that the notation sin²(θ) means (sin(θ))², and the same goes for cos. In other words you must take the sin first and then square. The reason for not placing the exponent at the end is to make it clear that it is not angle θ that is squared, rather it is the sin of θ that is squared.

Identity (4) is considered to be a basic simplification. We can also rearrange it as

1 − sin²(θ) = cos²(θ)

or as

1 − cos²(θ) = sin²(θ)

and these are also considered to be basic simplifications when they are used to replace two terms on the left with the one term on the right.

To prove identity (4) simply construct a right triangle with hypotenuse 1 and angle θ. Then the base and altitude are given by cos(θ) and sin(θ), and the original form of Pythagoras’ theorem, namely a² + b² = c², turns into identity (4).

The graph to the right illustrates Pythagoras’ theorem by showing how the height of the sin²(θ) curve (red) and the height of the cos²(θ) curve (blue) add to always equal 1 (black dashed line).

We can also write Pythagoras’ theorem in two other forms. If we divide both sides of identity (4) by cos²(θ) and simplify then it reads

1 + tan²(θ) = sec²(θ)

Alternatively, if we divide both sides of identity (4) by sin²(θ) and simplify then it reads

1 + cot²(θ) = csc²(θ)

The identities that we have discussed so far (identities 1 to 4) are really just simplifications that are always applied from left to right (i.e. by replacing the left-hand-side expression anywhere it appears by the right-hand-side expression.)

The rest of the identities (identities 5 to 9, discussed next) are stated so that the left-hand-side is the combined form of some expression and the right-hand-side is the broken apart form. They can be applied in either direction depending on what we want to accomplish:

From left to right: The purpose is to break apart a combined angle into two separate angles, or break one trigonometric function into two other ones, or break up any other trigonometric functions into sines and cosines. Going in this direction is easy. This technique is used to prove new trigonometric identities.
From right to left: The purpose is to combine two separate angles into one combined angle or to replace two terms or functions with one. Going in this direction is usually hard. This technique is used to create more compact expressions.

The tan identity

5.

This identity follows from the definitions of sin, cos and tan. To derive it simply divide the ratio for sin by the ratio for cos and simplify. The result is the ratio for tan:

The sum of angles identities

6a.

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

6b.

6c.

These identities describe how to break apart the 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 (6c) with the lower signs reads:

A common mistake is to ignore these identities and believe that, for example

sin(α + β) = sin(α)+ sin(β) WRONG!

This is not correct! It could only be correct if the sin curve was a straight line through the origin, instead of a wavy curve.

Here is an example that uses identity (6a) to describe what happens when a sin wave is shifted left by 30°.

We see that it becomes part sin wave and part cos wave. It is interesting to compare this result with identity (3a), which shows that shifting a sin wave left by 90° turns it completely into a cos wave.

It is possible (but more difficult) to use identity (6a) in the opposite direction and show how the sum of a sin and a cos can be expressed as a sin with a phase shift. Click here to see an example of that.

We will now prove that sin(α + β) = sin(α)·cos(β) + cos(α)·sin(β).

Proof: The picture shows a big yellow oblique triangle with two small right triangles inside it. Apply the sine law to the big triangle.

Use the fact that sin(γ) = h/c and then multiply (a + b) across to the other side.

Expand the right-hand-side and then use the two small triangles in the picture to replace each fraction by the appropriate trigonometric function.

We are done.

Identity (6a) with the upper sign, namely:

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

which we just proved, is probably the most important identity of the group. Every other identity in the group (and in fact almost any trigonometric identity) can be derived from it. For example:

The identity for sin(α − β) can be derived by writing it as sin(α + (−β)), applying identity (6a) and then using the negative angle identities, (2a) and (2b), to clean up.
The identity for cos(α + β) can be derived by using identity (3a) to write it as a sin with a phase shift:
sin((α + 90°) + β),
then applying identity (6a) and then using (3a) again to clean up.
The identity for tan(α + β) can be derived by dividing identity (6a) by (6b).

The double angle identities

7a.

sin(2α) = 2·sin(α)·cos(α)

7b.

cos(2α) = cos²(α) − sin²(α)

= 1 − 2 sin²(α)

= 2 cos²(α) − 1

7c.

These identities are special cases of identities (6a), (6b) and (6c) with the upper signs, which read:

Let β equal α and identities (7a), (7b), (7c) immediately pop out. Identity (7b) is written in 3 different forms. The second and third forms result from using Pythagoras’ theorem on the first form. They are the preferred forms because they only involve sin or only cos, and 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 (7b) 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:

8a.

8b.

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 (6a) and (6b) in various combinations then we get the so-called product identites:

9a.

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

9b.

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

9c. 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.

Examples

We finish this section with seven examples. The first three examples show how identites 1 to 9 can be used to prove new trigonometric identities.

The last four examples show how converting a trigonometric expression to another form leads to new insights that were not previously evident. The examples come from electrical technology where AC (alternating current) has the form of a sin wave.

Trigonometric identities are also used to help solve trigonometric equations. That topic is covered in the next section.

How to prove a trigonometric identity:

Proving an identity is different than solving an equation. Even though the identity contains an = sign you must not transpose a quantity from one side to the other because doing so changes the value of both sides. Instead you must use trigonometric identities to modify the left side or the right side or both sides until they are identical.
The first step is to use the reciprocal identities and the tan identity to replace tan, cot, sec, csc wherever they occur by sin and cos.
These identities produce fractions. Combine (add) the fractions.
Use the sum of angles identities or double angle identities to break apart any sums of angles or to replace double angles.
If you end up with a fraction on one side of the identity but not the other then multiply the non-fraction side by a UFOO to convert into a fraction. A UFOO is a fraction that equals 1 because it has equal numerator and denominator. Example 3 requires one.
When the two sides are identical the identity is proven.

Example 1. Prove the trigonometric identity

Solution: Follow these steps:

Details of the steps:

Use the tan identity to replace the tan function with sin/cos on the LHS (left-hand-side).
Do the same on the right-hand-side. Also since the tan and cot functions are reciprocals replace the cot function with cos/sin.
The right side is now a compound fraction (a fraction that contains more fractions). Combine the fractions in the numerator and the fractions in the denominator.
Use the invert and multiply rule to divide the fractions and simplify. The two sides are now identical so the identity is proven.

Example 2. There are two new features in this example:

There are several different angles involved.
We need to compare the two sides for guidance on what to do.

Prove the trigonometric identity

Solution: Follow these steps:

Details of the steps:

Use the reciprocal identities to replace the sec and csc functions with cos and sin.
Comparing the two sides we notice that the angles on the RHS are 4x and x. We can have the same angles on the LHS if we replace angle 5x by the sum of angles 4x + x.
Apply the sum of angles identity to sin(4x + x) in the numerator.
Break the single fraction into two fractions. The two sides are now identical so the identity is proven.

Example 3. The new feature here is that the right side is a fraction and the left side is not. We will have to make it a fraction.

Prove the trigonometric identity

Solution: Follow these steps:

Details of the steps:

Since the angle on the RHS is double the angle on the LHS, replace angle 2α on the RHS using the double angle identity for sin.
The RHS has a denominator sin(α) + cos(α). To get the same denominator on the LHS we apply a UFOO to it; that is, we multiply and divide the LHS by the same quantity, namely sin(α) + cos(α).
Expand the numerator.
Use Pythagoras’ theorem. The two sides are now identical so the identity is proven.

Example 4. Adding a sin wave and a cos wave of the same frequency produces another wave of the same frequency

(a) Prove the identity

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

(b) Use the identity to express the sum of the two sinusoidal waveforms,

y₁ = 3 sin(θ) and y₂ = 4 cos(θ),

as a single sin wave with a phase shift. Then plot all three waveforms in a graph.

Solution part (a): Because it is easier to break angles apart than it is to combine them we will work on the right-hand-side of this identity and convert it into the left-hand-side.

First notice that the pair of equations for C and φ describe the conversion of a vector from rectangular to polar coordinates. Referring to the figure to the right, the vector is (A, B) in rectangular coordinates and C∠φ in polar coordinates. For use in step 3 of the proof notice that the diagram also shows that

A = C cos(φ) and B = C sin(φ).

Here are the steps in the proof:

Details of the steps:

Use the sum of angles identity for sin to break apart angles θ and φ.
Expand.
Use the fact (mentioned above) that A = C cos(φ) and B = C sin(φ). The two sides are now identical so the identity is proven.

Solution part (b): We will now use the identity proven in part (a) to express the sum y = 3 sin(θ) + 4 cos(θ) as a single sin wave with a phase shift. Let A = 3 and B = 4 in the identity. This gives

so the sum of the waveforms y₁ = 3 sin(θ) and y₂ = 4 cos(θ) is

y = 5 sin(θ + 53.13°)

The graph to the right shows the separate waveforms in green and blue and the resultant waveform in red. (Click here to see how the red waveform can be graphed.)
Notes:

The significance of this example is that it shows that adding two sinusoidal waveforms of the same frequency results in another sinusoidal waveform of the same frequency but with a phase shift. Compare this result with the result found in the next example.
We could add any number of out-of-phase sin or cos waves of the same frequency to produce a single sin with a phase shift by following these steps:
1. Break each waveform into a sin and a cos by applying the identity from right to left (in the Algebra Coach click the Break apart trig button),
2. Add all the sines and add all the cosines,
3. Combine the remaining sin and cos by applying the identity from left to right (in the Algebra Coach click the Combine trig button).
One of the most important situations where one is concerned with out-of-phase waves is in an alternating current (AC) network. An example is the continent-wide electrical energy grid. Every generator that is connected to the grid must produce voltage that is in phase with all the other generators on the grid. Otherwise the voltages will tend to cancel.

Example 5. Waves with slightly different frequencies produce beats

(a) Prove the identity

(b) The left side of the identity can be considered to be the sum of two sinusoidal waveforms

with slightly different frequencies. Plot waveforms y₁ and y₂ on the same graph and compare them.

(c) Now plot their sum in another graph. This task would be very difficult were it not for the identity proven in part (a). The identity can be used to convert the sum of the waveforms into a single term. Show that this single term can be interpreted as a sin wave with a slowly varying amplitude. Use the single term side of the identity to make the plot.

Solution part (a): We are essentially proving the product identity (9b). The key idea in the proof is noticing that the angles 21t and 19t can be thought of a 20t ± 1t and then breaking the angles apart with a sum of angles identity.

Here are the steps in the proof:

Details of the steps:

The RHS has angles t and 20t. To get the same angles on the LHS write 21t as 20t + 1t and 19t as 20t − 1t.
Apply the sum of angles identity for sin to both terms. That identity says
sin(α ± β) = sin(α)·cos(β) ± cos(α)·sin(β)
In this case let α = 20t and β = 1t.
Simplify.

Solution part (b): Here is the graph of the two sinusoidal waveforms:

Notice that the waveform with angular velocity ω = 21 (the blue one) oscillates slightly faster than the waveform with ω = 19 (the green one). The blue one has 10.5 cycles in the time shown while the green one has 9.5 cycles; one cycle less. The result is that at certain times, for example at time t = 0 and t = 3.14, the waves are in phase (that is, peaking together), and at other times, for example at time t = 1.57, the waves are 180° out of phase, (that is, one is at a crest when the other is in a trough).

Solution part (c): It would be very difficult to add the waveforms graphically. All we could say for sure is that when they are in phase they add to give an amplitude of 1 and when they are 180° out of phase they cancel to give an amplitude of 0.

What is more informative is to make a plot of the other side of the identity, namely of

y = cos(t) · sin(20t).

This graph is plotted below in red. We see that the factor cos(t) in this product plays the role of a slowly varying amplitude for the rapidly varying oscillation sin(20t). That is, we have a sinusoidal wave sin(20t), but that at certain times it has a small amplitude and at other times it has a large amplitude.

To make this graph we first draw one cycle of the curve y = cos(t) (and its negative), both shown in gray. This is called the envelope and is used as a guide. Then we draw the curve y = sin(20t) inside the envelope. We draw 20 oscillations and stretch and squeeze them to fit inside the envelope.

Notes:

In acoustics the periodic constructive and destructive interference is known as beats. Beats can be quite annoying if they are due to, say, the twin engines of an airplane running at slightly different speeds. But they can be useful, for example, for checking if two guitar strings are in tune.
In electrical energy generation every generator that is connected to the electrical grid must produce voltage at the same frequency as all the other generators on the grid. Otherwise the voltages will tend to periodically cancel.

Example 6. Electric power dissipated by a resistor

There are two types of electric current in common use: direct current (DC) and alternating current (AC).

Direct current is current that is constant in time. It is typically produced by a battery. For this example we will consider a direct current of 0.7071 amperes:
i = 0.7071 amperes
Alternating current is current that flows back and forth in an electric circuit and is described by a sinusoidal function of time. It is typically produced by an electric generator. For this example we will use an alternating current with an amplitude of 1.0 amperes:
i = 1.0 sin(t) amperes

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

When an electric current flows through a resistor (such as a toaster or light bulb) it dissipates heat or light energy. The power (i.e. the rate at which the energy is dissipated) is given by the formula

p = i² R,

where p is the power in watts, i is the current in amperes, and R is the resistance of the resistor in ohms. If the current changes with time then the power must be calculated at each instant in time with this formula. We say that the formula gives the instantaneous power.

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

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

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 = 1000 Ω it 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 = 1000 Ω it gives exactly the same upside-down cosine wave shape for the instantaneous power. But when the average height of i², namely 0.5, is multiplied by R = 1000 Ω then this gives an average power p = 500 watts, as was to be shown.

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 half-angle identity (8a), modified to read .
Interpret the RHS of the identity. It is a sinusoidal waveform plus a DC component.
The last term is the DC component. It causes i² to have an average value of ½.

Notes:

This example showed that an alternating current with an amplitude of 1 ampere is as effective as a direct current of 0.7071 amperes in powering a resistor such as a lightbulb or toaster. This result can be generalized. The effective value (also called the RMS value) of any AC current is found by multiplying its amplitude by 0.7071.
In cases where the frequency of the alternating current is high the average value of the power is more relevant than the instantaneous power. An example is when a 60 Hz AC current goes through a kitchen toaster. The toaster element doesn’t have time to heat up and cool down as the current changes and just stays at one constant temperature. Another example is the same 60 Hz current flowing through a fluorescent light. In this case the light actually goes on and off as the current alternates, but the human eye can’t respond fast enough to see it. In both situations we are more interested in the average power than the instantaneous power.

Example 7. Electric power generated by a generator

In the electric circuit shown to the right the generator produces energy and the load consumes that energy. The power consumed by the load equals the power produced by the generator. This is the law of conservation of energy.

If the load is a resistor then the previous example shows how to calculate the power consumed by the resistor. In this example we make two changes:

We study the power that is produced by the generator.
We let the load contain an inductor or capacitor. An inductor is any device that creates a magnetic field when a current flows through it. An example is an electric motor. A capacitor is any device that creates an electric field when electric charge is stored in it. In reality every circuit has some capacitance and some inductance either incidentally or by design.

The main consequence of the load having inductance or capacitance is that the alternating current and voltage produced by the generator can be out-of-phase by up to 90°.

For example the picture shows a voltage out of phase with the current, with the voltage leading the current by 60°.

The power p delivered by the generator at any instant in time is given by the formula

p = i v,

where i is the current in amperes at that instant and v is the voltage of the generator in volts. Suppose that the generator produces the AC current

i = I₀ sin (ωt)

and that the load is such that the voltage is out of phase with the current by angle φ. (The phase shift φ could be positive or negative.)

v = V₀ sin (ωt+φ)

Since i and v change with time, so does p. But as in the previous example, rather than the instantaneous power, we are interested in the average power.

Problem: Find the average power delivered by the generator.

Solution: We follow the same steps as in the previous example. Substituting the given i and v into the power formula gives

We can change the product of trigonometric functions into a sum of trigonometric functions by using identity (9a) which says that sin(α)·sin(β) = ½[cos(α − β) − cos(α + β)]. If we make the replacements α→ωt+φ and β→ωt then identity (9a) reads

sin(ωt+φ)·sin(ωt) = ½[cos(φ) − cos(2ωt+φ)]

Substituting this into the formula for p gives

or expanding,

Because φ is a constant (remember it is the phase shift between v and i) the instantaneous power p is again a sinusoidal waveform with a DC component. The DC component gives the average power,

Some special loads.

Perfect resistors: i and v are in phase (φ = 0) so the above formula gives . Since V₀ = I₀ R this can also be written as
This agrees with the result we derived in the previous example, which was that p_avg = ( I_effective )² R where I_effective = 0.7071 I₀
Perfect inductors: φ = +90^o and perfect capacitors: φ = −90^o. In both cases the above formula gives
p_avg = 0.
How can this be? The explanation is that inductors and capacitors absorb energy for one half of each cycle as they build up their magnetic and electric fields and then release the energy back to the generator in the other half of the cycle when the fields collapse.

The following graph shows the instantaneous power (solid red curve) and the average power (dashed red line) that result from out-of-phase current i (blue curve) and voltage v (gray curve).

Note that at points a and b the instantaneous power is actually negative. The reason is that at those moments i and v have opposite signs and the explanation, as mentioned above, is that power is going back to the generator. Also note that if φ = 0 then the red curve reduces to the i² curve in the previous example.

Algebra Coach Exercises

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