Proofs of trigonometric identities

Definitions

Referring to the diagram at the right, the six trigonometric functions of θ are:

sin ⁡ θ = o p p o s i t e h y p o t e n u s e = a h cos ⁡ θ = a d j a c e n t h y p o t e n u s e = b h tan ⁡ θ = o p p o s i t e a d j a c e n t = a b cot ⁡ θ = a d j a c e n t o p p o s i t e = b a sec ⁡ θ = h y p o t e n u s e a d j a c e n t = h b csc ⁡ θ = h y p o t e n u s e o p p o s i t e = h a

Ratio identities

The following identities are trivial algebraic consequences of these definitions and the division identity.
They rely on multiplying or dividing the numerator and denominator of fractions by a variable. Ie,

a b = ( a h ) ( b h ) tan ⁡ θ = o p p o s i t e a d j a c e n t = ( o p p o s i t e h y p o t e n u s e ) ( a d j a c e n t h y p o t e n u s e ) = sin ⁡ θ cos ⁡ θ cot ⁡ θ = a d j a c e n t o p p o s i t e = ( a d j a c e n t a d j a c e n t ) ( o p p o s i t e a d j a c e n t ) = 1 tan ⁡ θ = cos ⁡ θ sin ⁡ θ sec ⁡ θ = 1 cos ⁡ θ = h y p o t e n u s e a d j a c e n t csc ⁡ θ = 1 sin ⁡ θ = h y p o t e n u s e o p p o s i t e tan ⁡ θ = o p p o s i t e a d j a c e n t = ( o p p o s i t e × h y p o t e n u s e o p p o s i t e × a d j a c e n t ) ( a d j a c e n t × h y p o t e n u s e o p p o s i t e × a d j a c e n t ) = ( h y p o t e n u s e a d j a c e n t ) ( h y p o t e n u s e o p p o s i t e ) = sec ⁡ θ csc ⁡ θ

Or

tan ⁡ θ = sin ⁡ θ cos ⁡ θ = ( 1 csc ⁡ θ ) ( 1 sec ⁡ θ ) = ( csc ⁡ θ sec ⁡ θ csc ⁡ θ ) ( csc ⁡ θ sec ⁡ θ sec ⁡ θ ) = sec ⁡ θ csc ⁡ θ cot ⁡ θ = csc ⁡ θ sec ⁡ θ

Complementary angle identities

Two angles whose sum is π/2 radians (90 degrees) are complementary. In the diagram, the angles at vertices A and B are complementary, so we can exchange a and b, and change θ to π/2 − θ, obtaining:

sin ⁡ ( π / 2 − θ ) = cos ⁡ θ cos ⁡ ( π / 2 − θ ) = sin ⁡ θ tan ⁡ ( π / 2 − θ ) = cot ⁡ θ cot ⁡ ( π / 2 − θ ) = tan ⁡ θ sec ⁡ ( π / 2 − θ ) = csc ⁡ θ csc ⁡ ( π / 2 − θ ) = sec ⁡ θ

Pythagorean identities

Identity 1:

sin 2 ⁡ ( x ) + cos 2 ⁡ ( x ) = 1

The following two results follow from this and the ratio identities. To obtain the first, divide both sides of sin 2 ⁡ ( x ) + cos 2 ⁡ ( x ) = 1 by cos 2 ⁡ ( x ) ; for the second, divide by sin 2 ⁡ ( x ) .

tan 2 ⁡ ( x ) + 1   = sec 2 ⁡ ( x ) 1   + cot 2 ⁡ ( x ) = csc 2 ⁡ ( x )

Similarly

1   + cot 2 ⁡ ( x ) = csc 2 ⁡ ( x ) csc 2 ⁡ ( x ) − cot 2 ⁡ ( x ) = 1  

Identity 2:

The following accounts for all three reciprocal functions.

csc 2 ⁡ ( x ) + sec 2 ⁡ ( x ) − cot 2 ⁡ ( x ) = 2   + tan 2 ⁡ ( x )

Proof 2:

Refer to the triangle diagram above. Note that a 2 + b 2 = h 2 by Pythagorean theorem.

csc 2 ⁡ ( x ) + sec 2 ⁡ ( x ) = h 2 a 2 + h 2 b 2 = a 2 + b 2 a 2 + a 2 + b 2 b 2 = 2   + b 2 a 2 + a 2 b 2

Substituting with appropriate functions -

2   + b 2 a 2 + a 2 b 2 = 2   + tan 2 ⁡ ( x ) + cot 2 ⁡ ( x )

Rearranging gives:

csc 2 ⁡ ( x ) + sec 2 ⁡ ( x ) − cot 2 ⁡ ( x ) = 2   + tan 2 ⁡ ( x )

Sine

Draw a horizontal line (the x-axis); mark an origin O. Draw a line from O at an angle α above the horizontal line and a second line at an angle β above that; the angle between the second line and the x-axis is α + β .

Place P on the line defined by α + β at a unit distance from the origin.

Let PQ be a line perpendicular to line defined by angle α , drawn from point Q on this line to point P. ∴ OQP is a right angle.

Let QA be a perpendicular from point A on the x-axis to Q and PB be a perpendicular from point B on the x-axis to P. ∴ OAQ and OBP are right angles.

Draw R on PB so that QR is parallel to the x-axis.

Now angle R P Q = α (because O Q A = π 2 − α , making R Q O = α , R Q P = π 2 − α , and finally R P Q = α )

R P Q = π 2 − R Q P = π 2 − ( π 2 − R Q O ) = R Q O = α O P = 1 P Q = sin ⁡ β O Q = cos ⁡ β A Q O Q = sin ⁡ α , so A Q = sin ⁡ α cos ⁡ β P R P Q = cos ⁡ α , so P R = cos ⁡ α sin ⁡ β sin ⁡ ( α + β ) = P B = R B + P R = A Q + P R = sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β

By substituting − β for β and using Symmetry, we also get:

sin ⁡ ( α − β ) = sin ⁡ α cos ⁡ ( − β ) + cos ⁡ α sin ⁡ ( − β ) sin ⁡ ( α − β ) = sin ⁡ α cos ⁡ β − cos ⁡ α sin ⁡ β

Another rigorous proof, and much easier, can be given by using Euler's formula, known from complex analysis. Euler's formula is:

e i φ = cos ⁡ φ + i sin ⁡ φ

It follows that for angles α and β we have:

e i ( α + β ) = cos ⁡ ( α + β ) + i sin ⁡ ( α + β )

Also using the following properties of exponential functions:

e i ( α + β ) = e i α e i β = ( cos ⁡ α + i sin ⁡ α ) ( cos ⁡ β + i sin ⁡ β )

Evaluating the product:

e i ( α + β ) = ( cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ β ) + i ( sin ⁡ α cos ⁡ β + sin ⁡ β cos ⁡ α )

Equating real and imaginary parts:

cos ⁡ ( α + β ) = cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ β sin ⁡ ( α + β ) = sin ⁡ α cos ⁡ β + sin ⁡ β cos ⁡ α

Cosine

Using the figure above,

O P = 1 P Q = sin ⁡ β O Q = cos ⁡ β O A O Q = cos ⁡ α , so O A = cos ⁡ α cos ⁡ β R Q P Q = sin ⁡ α , so R Q = sin ⁡ α sin ⁡ β cos ⁡ ( α + β ) = O B = O A − B A = O A − R Q = cos ⁡ α cos ⁡ β   − sin ⁡ α sin ⁡ β

By substituting − β for β and using Symmetry, we also get:

cos ⁡ ( α − β ) = cos ⁡ α cos ⁡ ( − β ) − sin ⁡ α sin ⁡ ( − β ) , cos ⁡ ( α − β ) = cos ⁡ α cos ⁡ β + sin ⁡ α sin ⁡ β

Also, using the complementary angle formulae,

cos ⁡ ( α + β ) = sin ⁡ ( π / 2 − ( α + β ) ) = sin ⁡ ( ( π / 2 − α ) − β ) = sin ⁡ ( π / 2 − α ) cos ⁡ β − cos ⁡ ( π / 2 − α ) sin ⁡ β = cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ β

Tangent and cotangent

From the sine and cosine formulae, we get

tan ⁡ ( α + β ) = sin ⁡ ( α + β ) cos ⁡ ( α + β ) = sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ β

Dividing both numerator and denominator by cos ⁡ α cos ⁡ β , we get

tan ⁡ ( α + β ) = tan ⁡ α + tan ⁡ β 1 − tan ⁡ α tan ⁡ β

Subtracting β from α , using tan ⁡ ( − β ) = − tan ⁡ β ,

tan ⁡ ( α − β ) = tan ⁡ α + tan ⁡ ( − β ) 1 − tan ⁡ α tan ⁡ ( − β ) = tan ⁡ α − tan ⁡ β 1 + tan ⁡ α tan ⁡ β

Similarly from the sine and cosine formulae, we get

cot ⁡ ( α + β ) = cos ⁡ ( α + β ) sin ⁡ ( α + β ) = cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ β sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β

Then by dividing both numerator and denominator by sin ⁡ α sin ⁡ β , we get

cot ⁡ ( α + β ) = cot ⁡ α cot ⁡ β − 1 cot ⁡ α + cot ⁡ β

Or, using cot ⁡ θ = 1 tan ⁡ θ ,

cot ⁡ ( α + β ) = 1 − tan ⁡ α tan ⁡ β tan ⁡ α + tan ⁡ β = 1 tan ⁡ α tan ⁡ β − 1 1 tan ⁡ α + 1 tan ⁡ β = cot ⁡ α cot ⁡ β − 1 cot ⁡ α + cot ⁡ β

Using cot ⁡ ( − β ) = − cot ⁡ β ,

cot ⁡ ( α − β ) = cot ⁡ α cot ⁡ ( − β ) − 1 cot ⁡ α + cot ⁡ ( − β ) = cot ⁡ α cot ⁡ β + 1 cot ⁡ β − cot ⁡ α

Double-angle identities

From the angle sum identities, we get

sin ⁡ ( 2 θ ) = 2 sin ⁡ θ cos ⁡ θ

and

cos ⁡ ( 2 θ ) = cos 2 ⁡ θ − sin 2 ⁡ θ

The Pythagorean identities give the two alternative forms for the latter of these:

cos ⁡ ( 2 θ ) = 2 cos 2 ⁡ θ − 1 cos ⁡ ( 2 θ ) = 1 − 2 sin 2 ⁡ θ

The angle sum identities also give

tan ⁡ ( 2 θ ) = 2 tan ⁡ θ 1 − tan 2 ⁡ θ = 2 cot ⁡ θ − tan ⁡ θ cot ⁡ ( 2 θ ) = cot 2 ⁡ θ − 1 2 cot ⁡ θ = cot ⁡ θ − tan ⁡ θ 2

It can also be proved using Euler's formula

e i φ = cos ⁡ φ + i sin ⁡ φ

Squaring both sides yields

e i 2 φ = ( cos ⁡ φ + i sin ⁡ φ ) 2

But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields

e i 2 φ = cos ⁡ 2 φ + i sin ⁡ 2 φ

It follows that

( cos ⁡ φ + i sin ⁡ φ ) 2 = cos ⁡ 2 φ + i sin ⁡ 2 φ .

Expanding the square and simplifying on the left hand side of the equation gives

i ( 2 sin ⁡ φ cos ⁡ φ ) + cos 2 ⁡ φ − sin 2 ⁡ φ   = cos ⁡ 2 φ + i sin ⁡ 2 φ .

Because the imaginary and real parts have to be the same, we are left with the original identities

cos 2 ⁡ φ − sin 2 ⁡ φ   = cos ⁡ 2 φ ,

and also

2 sin ⁡ φ cos ⁡ φ = sin ⁡ 2 φ .

Half-angle identities

The two identities giving the alternative forms for cos 2θ lead to the following equations:

cos ⁡ θ 2 = ± 1 + cos ⁡ θ 2 , sin ⁡ θ 2 = ± 1 − cos ⁡ θ 2 .

The sign of the square root needs to be chosen properly—note that if π is added to θ, the quantities inside the square roots are unchanged, but the left-hand-sides of the equations change sign. Therefore, the correct sign to use depends on the value of θ.

For the tan function, the equation is:

tan ⁡ θ 2 = ± 1 − cos ⁡ θ 1 + cos ⁡ θ .

Then multiplying the numerator and denominator inside the square root by (1 + cos θ) and using Pythagorean identities leads to:

tan ⁡ θ 2 = sin ⁡ θ 1 + cos ⁡ θ .

Also, if the numerator and denominator are both multiplied by (1 - cos θ), the result is:

tan ⁡ θ 2 = 1 − cos ⁡ θ sin ⁡ θ .

This also gives:

tan ⁡ θ 2 = csc ⁡ θ − cot ⁡ θ .

Similar manipulations for the cot function give:

cot ⁡ θ 2 = ± 1 + cos ⁡ θ 1 − cos ⁡ θ = 1 + cos ⁡ θ sin ⁡ θ = sin ⁡ θ 1 − cos ⁡ θ = csc ⁡ θ + cot ⁡ θ .

Miscellaneous -- the triple tangent identity

If ψ + θ + ϕ = π = half circle (for example, ψ , θ and ϕ are the angles of a triangle),

tan ⁡ ( ψ ) + tan ⁡ ( θ ) + tan ⁡ ( ϕ ) = tan ⁡ ( ψ ) tan ⁡ ( θ ) tan ⁡ ( ϕ ) .

Proof:

ψ = π − θ − ϕ tan ⁡ ( ψ ) = tan ⁡ ( π − θ − ϕ ) = − tan ⁡ ( θ + ϕ ) = − tan ⁡ θ − tan ⁡ ϕ 1 − tan ⁡ θ tan ⁡ ϕ = tan ⁡ θ + tan ⁡ ϕ tan ⁡ θ tan ⁡ ϕ − 1 ( tan ⁡ θ tan ⁡ ϕ − 1 ) tan ⁡ ψ = tan ⁡ θ + tan ⁡ ϕ tan ⁡ ψ tan ⁡ θ tan ⁡ ϕ − tan ⁡ ψ = tan ⁡ θ + tan ⁡ ϕ tan ⁡ ψ tan ⁡ θ tan ⁡ ϕ = tan ⁡ ψ + tan ⁡ θ + tan ⁡ ϕ

Miscellaneous -- the triple cotangent identity

If ψ + θ + ϕ = π 2 = quarter circle,

cot ⁡ ( ψ ) + cot ⁡ ( θ ) + cot ⁡ ( ϕ ) = cot ⁡ ( ψ ) cot ⁡ ( θ ) cot ⁡ ( ϕ ) .

Proof:

Replace each of ψ , θ , and ϕ with their complementary angles, so cotangents turn into tangents and vice versa.

Given

ψ + θ + ϕ = π 2 ∴ ( π 2 − ψ ) + ( π 2 − θ ) + ( π 2 − ϕ ) = 3 π 2 − ( ψ + θ + ϕ ) = 3 π 2 − π 2 = π

so the result follows from the triple tangent identity.

Prosthaphaeresis identities

sin ⁡ θ ± sin ⁡ ϕ = 2 sin ⁡ ( θ ± ϕ 2 ) cos ⁡ ( θ ∓ ϕ 2 )

cos ⁡ θ + cos ⁡ ϕ = 2 cos ⁡ ( θ + ϕ 2 ) cos ⁡ ( θ − ϕ 2 )

cos ⁡ θ − cos ⁡ ϕ = − 2 sin ⁡ ( θ + ϕ 2 ) sin ⁡ ( θ − ϕ 2 )

Proof of sine identities

First, start with the sum-angle identities:

sin ⁡ ( α + β ) = sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β sin ⁡ ( α − β ) = sin ⁡ α cos ⁡ β − cos ⁡ α sin ⁡ β

By adding these together,

sin ⁡ ( α + β ) + sin ⁡ ( α − β ) = sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β + sin ⁡ α cos ⁡ β − cos ⁡ α sin ⁡ β = 2 sin ⁡ α cos ⁡ β

Similarly, by subtracting the two sum-angle identities,

sin ⁡ ( α + β ) − sin ⁡ ( α − β ) = sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β − sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β = 2 cos ⁡ α sin ⁡ β

Let α + β = θ and α − β = ϕ ,

∴ α = θ + ϕ 2 and β = θ − ϕ 2

Substitute θ and ϕ

sin ⁡ θ + sin ⁡ ϕ = 2 sin ⁡ ( θ + ϕ 2 ) cos ⁡ ( θ − ϕ 2 ) sin ⁡ θ − sin ⁡ ϕ = 2 cos ⁡ ( θ + ϕ 2 ) sin ⁡ ( θ − ϕ 2 ) = 2 sin ⁡ ( θ − ϕ 2 ) cos ⁡ ( θ + ϕ 2 )

Therefore,

sin ⁡ θ ± sin ⁡ ϕ = 2 sin ⁡ ( θ ± ϕ 2 ) cos ⁡ ( θ ∓ ϕ 2 )

Proof of cosine identities

Similarly for cosine, start with the sum-angle identities:

cos ⁡ ( α + β ) = cos ⁡ α cos ⁡ β   − sin ⁡ α sin ⁡ β cos ⁡ ( α − β ) = cos ⁡ α cos ⁡ β + sin ⁡ α sin ⁡ β

Again, by adding and subtracting

cos ⁡ ( α + β ) + cos ⁡ ( α − β ) = cos ⁡ α cos ⁡ β   − sin ⁡ α sin ⁡ β + cos ⁡ α cos ⁡ β + sin ⁡ α sin ⁡ β = 2 cos ⁡ α cos ⁡ β   cos ⁡ ( α + β ) − cos ⁡ ( α − β ) = cos ⁡ α cos ⁡ β   − sin ⁡ α sin ⁡ β − cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ β = − 2 sin ⁡ α sin ⁡ β

Substitute θ and ϕ as before,

cos ⁡ θ + cos ⁡ ϕ = 2 cos ⁡ ( θ + ϕ 2 ) cos ⁡ ( θ − ϕ 2 ) cos ⁡ θ − cos ⁡ ϕ = − 2 sin ⁡ ( θ + ϕ 2 ) sin ⁡ ( θ − ϕ 2 )

Inequalities

The figure at the right shows a sector of a circle with radius 1. The sector is θ/(2π) of the whole circle, so its area is θ/2.

O A = O D = 1 A B = sin ⁡ θ C D = tan ⁡ θ

The area of triangle OAD is AB/2, or sinθ/2. The area of triangle OCD is CD/2, or tanθ/2.

Since triangle OAD lies completely inside the sector, which in turn lies completely inside triangle OCD, we have

sin ⁡ θ < θ < tan ⁡ θ

This geometric argument applies if 0<θ<π/2. It relies on definitions of arc length and area, which act as assumptions, so it is rather a condition imposed in construction of trigonometric functions than a provable property. For the sine function, we can handle other values. If θ>π/2, then θ>1. But sinθ≤1 (because of the Pythagorean identity), so sinθ<θ. So we have

sin ⁡ θ θ < 1       i f       0 < θ

For negative values of θ we have, by symmetry of the sine function

sin ⁡ θ θ = sin ⁡ ( − θ ) − θ < 1

Hence

sin ⁡ θ θ < 1       i f       θ ≠ 0 tan ⁡ θ θ > 1       i f       0 < θ < π 2

Preliminaries

lim θ → 0 sin ⁡ θ = 0 lim θ → 0 cos ⁡ θ = 1

Sine and angle ratio identity

lim θ → 0 sin ⁡ θ θ = 1

Proof: From the previous inequalities, we have, for small angles

sin ⁡ θ < θ < tan ⁡ θ ,

Therefore,

sin ⁡ θ θ < 1 < tan ⁡ θ θ ,

Consider the right-hand inequality. Since

tan ⁡ θ = sin ⁡ θ cos ⁡ θ ∴ 1 < sin ⁡ θ θ cos ⁡ θ

Multiply through by cos ⁡ θ

cos ⁡ θ < sin ⁡ θ θ

Combining with the left-hand inequality:

cos ⁡ θ < sin ⁡ θ θ < 1

Taking cos ⁡ θ to the limit as θ → 0

lim θ → 0 cos ⁡ θ = 1

Therefore,

lim θ → 0 sin ⁡ θ θ = 1

Cosine and angle ratio identity

lim θ → 0 1 − cos ⁡ θ θ = 0

Proof:

1 − cos ⁡ θ θ = 1 − cos 2 ⁡ θ θ ( 1 + cos ⁡ θ ) = sin 2 ⁡ θ θ ( 1 + cos ⁡ θ ) = ( sin ⁡ θ θ ) × sin ⁡ θ × ( 1 1 + cos ⁡ θ )

The limits of those three quantities are 1, 0, and 1/2, so the resultant limit is zero.

Cosine and square of angle ratio identity

lim θ → 0 1 − cos ⁡ θ θ 2 = 1 2

Proof:

As in the preceding proof,

1 − cos ⁡ θ θ 2 = sin ⁡ θ θ × sin ⁡ θ θ × 1 1 + cos ⁡ θ .

The limits of those three quantities are 1, 1, and 1/2, so the resultant limit is 1/2.

Proof of Compositions of trig and inverse trig functions

All these functions follow from the Pythagorean trigonometric identity. We can prove for instance the function

sin ⁡ [ arctan ⁡ ( x ) ] = x 1 + x 2

Proof:

We start from

sin 2 ⁡ θ + cos 2 ⁡ θ = 1

Then we divide this equation by cos 2 ⁡ θ

cos 2 ⁡ θ = 1 tan 2 ⁡ θ + 1

Then use the substitution θ = arctan ⁡ ( x ) , also use the Pythagorean trigonometric identity:

1 − sin 2 ⁡ [ arctan ⁡ ( x ) ] = 1 tan 2 ⁡ [ arctan ⁡ ( x ) ] + 1

Then we use the identity tan ⁡ [ arctan ⁡ ( x ) ] ≡ x

sin ⁡ [ arctan ⁡ ( x ) ] = x x 2 + 1