Proofs of trigonometric identities

Proofs of trigonometric identities

Proofs of trigonometric identities are used to show relations between trigonometric functions. This article will list trigonometric identities and prove them.

Elementary trigonometric identities

Definitions

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

: sin heta = frac {mathrm{opposite{mathrm{hypotenuse = frac {a}{h}

: cos heta = frac {mathrm{adjacent{mathrm{hypotenuse = frac {b}{h}

: an heta = frac {mathrm{opposite{mathrm{adjacent = frac {a}{b}

: cot heta = frac {mathrm{adjacent{mathrm{opposite = frac {b}{a}

: sec heta = frac {mathrm{hypotenuse{mathrm{adjacent = frac {h}{b}

: csc heta = frac {mathrm{hypotenuse{mathrm{opposite = frac {h}{a}

Ratio identities

The following identities are trivial algebraic consequences of these definitions and the division identity

: frac {a}{b}= frac {left(frac {a}{c} ight)} {left(frac {b}{c} ight) }.

: an heta= frac mathrm{opposite}mathrm{adjacent}= frac { left( frac mathrm{opposite} mathrm{hypotenuse} ight) } { left( frac mathrm{adjacent} mathrm{hypotenuse} ight) }= frac {sin heta} {cos heta}.

: cot heta = frac {cos heta}{sin heta}.

: cot heta =frac mathrm{adjacent} mathrm{opposite}= frac { left( frac mathrm{adjacent}mathrm{adjacent} ight) } { left( frac mathrm{opposite}mathrm{adjacent} ight) } = frac {1}{ an heta}.

: sec heta = frac {1}{cos heta}

: csc heta = frac {1}{sin heta}

: an heta = frac mathrm{opposite}mathrm{adjacent}= frac {left(frac {mathrm{opposite} imes mathrm{hypotenuse{mathrm{opposite} imes mathrm{adjacent ight) } { left( frac {mathrm{adjacent} imes mathrm{hypotenuse {mathrm{opposite} imes mathrm{adjacent} } ight) }= frac { left( frac mathrm{hypotenuse} mathrm{adjacent} ight)} { left( frac mathrm{hypotenuse} mathrm{opposite} ight) }= frac {sec heta}{csc heta}.

: cot heta = frac {csc heta}{sec heta}.

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:

: sinleft( pi/2- heta ight) = cos heta

: cosleft( pi/2- heta ight) = sin heta

: anleft( pi/2- heta ight) = cot heta

: cotleft( pi/2- heta ight) = an heta

: secleft( pi/2- heta ight) = csc heta

: cscleft( pi/2- heta ight) = sec heta

Pythagorean identities

:sin^2(x) + cos^2(x) = 1,

Proof 1:

Refer to the triangle diagram above. Note that a^2+b^2=h^2 by Pythagorean theorem.:sin^2(x) + cos^2(x) = frac{a^2}{h^2} + frac{b^2}{h^2} = frac{a^2+b^2}{h^2} = frac{h^2}{h^2} = 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).

Proof 2:

:cos^2 x + sin^2 x = cos(x - x) = cos 0 = 1,

: an^2 x + 1 = sec^2 x !

:cot^2 x + 1 = csc^2 x !

Proof 3:

Differentiating the left-hand side of the identity yields:

:2 sin x cos x - 2 sin x cos x = 0

Integrating this shows that the original identity is equal to a constant, and this constant can be found by plugging in any arbitrary value of x.

Angle sum identities

ine

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

Let PQ be a perpendicular from P to the line defined by the angle α.OQP is a right angle.

Let QA be a perpendicular from Q to the x axis, and PB be a perpendicular from P to the x axis.OAQ is a right angle.

Draw QR parallel to the "x"-axis.Now angle RPQ = α (because RPQ = frac{pi}{2} - RQP = frac{pi}{2} - ( frac{pi}{2} - RQO) = RQO = alpha)

:OP = 1,

:PQ = sin eta,

:OQ = cos eta,

:frac{AQ}{OQ} = sin alpha,, so AQ = sin alpha cos eta,

:frac{PR}{PQ} = cos alpha,, so PR = cos alpha sin eta,

:sin (alpha + eta) = PB = RB+PR = AQ+PR = sin alpha cos eta + cos alpha sin eta,

From the negative argument formulae, we also get

:sin (alpha - eta) = sin alpha cos eta - cos alpha sin eta,

Another simple "proof" can be given using Euler's formula known from complex analysis:Euler's formula is:

:e^{ivarphi}=cos varphi +i sin varphi

Although it is more precise to say that Euler's formula entails the trignometric identities, it follows that for angles alpha and eta we have:

:e^{i (alpha + eta)} = cos (alpha +eta) + i sin(alpha +eta)

Also using the following properties of exponential functions:

:e^{i(alpha + eta)} = e^{i alpha} e^{ieta}= (cos alpha +i sin alpha) (cos eta + i sin eta)

Evaluating the product::e^{i(alpha + eta)} = (cos alpha cos eta - sin alpha sin eta)+i(sin alpha cos eta + sin eta cos alpha)

This will only be equal to the previous expression we got, if the imaginary and real parts are equal respectively. Hence we get::cos (alpha +eta)=cos alpha cos eta - sin alpha sin eta

:sin (alpha +eta)=sin alpha cos eta + sin eta cos alpha

Cosine

Using the figure above,

:OP = 1,

:PQ = sin eta,

:OQ = cos eta,

:frac{OA}{OQ} = cos alpha,, so OA = cos alpha cos eta,

:frac{RQ}{PQ} = sin alpha,, so RQ = sin alpha sin eta,

:cos (alpha + eta) = OB = OA-BA = OA-RQ = cos alpha cos eta - sin alpha sin eta,

From the negative argument formulae, we also get

:cos (alpha - eta) = cos alpha cos eta + sin alpha sin eta,

Also, using the complementary angle formulae,

:cos (alpha + eta) = sinleft( pi/2-(alpha + eta) ight) = sinleft( (pi/2-alpha) - eta ight),::= sinleft( pi/2-alpha ight) cos eta - cosleft( pi/2-alpha ight) sin eta,::= cos alpha cos eta - sin alpha sin eta,

From the negative argument formulae, we also get

:cos (alpha - eta) = cos alpha cos eta + sin alpha sin eta,

Tangent and cotangent

From the sine and cosine formulae, we get: an (alpha + eta) = frac{sin (alpha + eta)}{cos (alpha + eta)},

:= frac{sin alpha cos eta + cos alpha sin eta}{cos alpha cos eta - sin alpha sin eta},

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

: an (alpha + eta) = frac{ an alpha + an eta}{1 - an alpha an eta},

: an (alpha - eta) = frac{ an alpha - an eta}{1 + an alpha an eta},

Similarly (using a division by sin α sin β), we get

:cot (alpha + eta) = frac{cot alpha cot eta - 1}{cot alpha + cot eta},

:cot (alpha - eta) = frac{cot alpha cot eta + 1}{cot eta - cot alpha},

Double-angle identities

From the angle sum identities, we get

:sin (2 heta) = 2 sin heta cos heta,

and

:cos (2 heta) = cos^2 heta - sin^2 heta,

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

:cos (2 heta) = 2 cos^2 heta - 1,

:cos (2 heta) = 1 - 2 sin^2 heta,

The angle sum identities also give

: an (2 heta) = frac{2 an heta}{1 - an^2 heta} = frac{2}{cot heta - an heta},

:cot (2 heta) = frac{cot^2 heta - 1}{2 cot heta} = frac{cot heta - an heta}{2},

Half-angle identities

The two identities giving alternative forms for cos 2θ give these:

:cos frac{ heta}{2} = pm, sqrtfrac{1 + cos heta}{2},,

:sin frac{ heta}{2} = pm, sqrtfrac{1 - cos heta}{2}.,

One must choose the sign of the square root properly—note that if 2π 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, we have

: an frac{ heta}{2} = pm, sqrtfrac{1 - cos heta}{1 + cos heta}.,

If we multiply the numerator and denominator inside the square root by (1 + cos θ), and do a little manipulation using the Pythagorean identities, we get

: an frac{ heta}{2} = frac{sin heta}{1 + cos heta}.,

If instead we multiply the numerator and denominator by (1 - cos θ), we get

: an frac{ heta}{2} = frac{1 - cos heta}{sin heta}.,

This also gives

: an frac{ heta}{2} = csc heta - cot heta.,

Similar manipulations for the cot function give

:cot frac{ heta}{2} = pm, sqrtfrac{1 + cos heta}{1 - cos heta} = frac{1 + cos heta}{sin heta} = frac{sin heta}{1 - cos heta} = csc heta + cot heta.,

Prosthaphaeresis identities

* sin heta pm sin gamma = 2 sin frac{ hetapm gamma}2 cos frac{ hetamp gamma}2
* cos heta + cos gamma = 2 cos frac{ heta+gamma}2 cos frac{ heta-gamma}2
* cos heta - cos gamma = -2 sin frac{ heta+gamma}2 sin frac{ heta-gamma}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.

:OA = OD = 1,

:AB = sin heta,

:CD = an heta,

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 heta < heta < an heta,

This geometric argument applies if 0<&theta;<&pi;/2. For the sine function, we can handle other values. If &theta;>&pi;/2, then &theta;>1. But sin&theta;&le;1 (because of the Pythagorean identity), so sin&theta;<&theta;. So we have

:frac{sin heta}{ heta} < 1 mathrm{if} 0 < heta,

For negative values of &theta; we have, by symmetry of the sine function

:frac{sin heta}{ heta} = frac{sin (- heta)}{- heta} < 1,

Hence

:frac{sin heta}{ heta} < 1 mathrm{if} heta e 0,

:frac{ an heta}{ heta} > 1 mathrm{if} 0 < heta < frac{pi}{2},

Identities involving calculus

Preliminaries

:lim_{ heta o 0}{sin heta} = 0,

:lim_{ heta o 0}{cos heta} = 1,

These can be seen from looking at the diagrams.

ine and angle ratio identity

:lim_{ heta o 0}{frac{sin heta}{ heta = 1

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

:sin heta < heta < an heta,, so

:frac{sin heta}{ heta} < 1 < frac{ an heta}{ heta},, so

:frac{sin heta}{ heta cos heta} > 1,, or

:frac{sin heta}{ heta} > cos heta,, so

:cos heta < frac{sin heta}{ heta} < 1,, but

:lim_{ heta o 0}{cos heta} = 1,, so

:lim_{ heta o 0}{frac{sin heta}{ heta = 1

Cosine and angle ratio identity

:lim_{ heta o 0}frac{1 - cos heta}{ heta} = 0,

Proof:

:frac{1 - cos heta}{ heta} = frac{1 - cos^2 heta}{ heta (1 + cos heta)},

:= frac{sin^2 heta}{ heta (1 + cos heta)},

:= frac{sin heta}{ heta} imes sin heta imes frac{1}{1 + cos heta}.,

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_{ heta o 0}frac{1 - cos heta}{ heta^2} = frac{1}{2}

Proof:

As in the preceding proof,

:frac{1 - cos heta}{ heta^2} = frac{sin heta}{ heta} imes frac{sin heta}{ heta} imes frac{1}{1 + cos heta}.,

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

Pedagogical note

Most of the proofs on this page are geometric in nature, and use the geometric definitions of the functions. These proofs are intended to be accessible and familiar to students without training in calculus.

While the theorems and proofs of plane Euclidean geometry have a revered place in the history of mathematics, going back 2300 years to Euclid's Elements, this treatment of trigonometry is not considered rigorous by modern standards. Note in particular that the inequalities sin&theta; < &theta; < tan&theta; involve looking at a picture and observing that a triangle lies inside or outside a circular sector. Furthermore, it requires a knowledge of the formula π r2 for the area of a circle. While this doesn't require calculus "per se", the elementary derivation of this formula usually involves "calculus-like" thinking&mdash;the circle is approximated by triangles and it is observed that "in the limit" those triangles approach π r2.

A rigorous treatment of the trigonometric identities would begin with a rigorous definition of the functions, as power series, as complex exponentials, or through integrals. For example, the arctangent function can be defined as:arctan x = int_0^x frac{dt}{1+t^2},.After proving the other identities, one could prove inequalities such as:frac{sin x}{x} < 1,either through the power series definition or through the mean value theorem.

A rigorous derivation of the functions and theorems of trigonometry may be found in the appendix of Whittaker and Watson.

References

* E. T. Whittaker and G. N. Watson. "A course of modern analysis", Cambridge University Press, 1952


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