Proofs of trigonometric identities

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

[edit] Elementary trigonometric identities

[edit] Definitions

Trigonometric functions specify the relationships between side lengths and interior angles of a right triangle. For example, the sine of angle θ is defined as being the length of the opposite side divided by the length of the hypotenuse.

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

$\sin \theta = \frac {\mathrm{opposite}}{\mathrm{hypotenuse}} = \frac {a}{h}$

$\cos \theta = \frac {\mathrm{adjacent}}{\mathrm{hypotenuse}} = \frac {b}{h}$

$\tan \theta = \frac {\mathrm{opposite}}{\mathrm{adjacent}} = \frac {a}{b}$

$\cot \theta = \frac {\mathrm{adjacent}}{\mathrm{opposite}} = \frac {b}{a}$

$\sec \theta = \frac {\mathrm{hypotenuse}}{\mathrm{adjacent}} = \frac {h}{b}$

$\csc \theta = \frac {\mathrm{hypotenuse}}{\mathrm{opposite}} = \frac {h}{a}$

[edit] Ratio identities

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

$\frac {a}{b}= \frac {\left(\frac {a}{c}\right)} {\left(\frac {b}{c}\right) }.$

$\tan \theta = \frac \mathrm{opposite}\mathrm{adjacent} = \frac { \left( \frac \mathrm{opposite} \mathrm{hypotenuse} \right) } { \left( \frac \mathrm{adjacent} \mathrm{hypotenuse} \right) } = \frac {\sin \theta} {\cos \theta}.$

$\cot \theta = \frac {\cos \theta}{\sin \theta}.$

$\cot \theta =\frac \mathrm{adjacent} \mathrm{opposite} = \frac { \left( \frac \mathrm{adjacent}\mathrm{adjacent} \right) } { \left( \frac \mathrm{opposite}\mathrm{adjacent} \right) } = \frac {1}{\tan \theta}.$

$\sec \theta = \frac {1}{\cos \theta}$

$\csc \theta = \frac {1}{\sin \theta}$

$\tan \theta = \frac \mathrm{opposite}\mathrm{adjacent} = \frac {\left(\frac {\mathrm{opposite} \times \mathrm{hypotenuse}}{\mathrm{opposite} \times \mathrm{adjacent}} \right) } { \left( \frac {\mathrm{adjacent} \times \mathrm{hypotenuse}} {\mathrm{opposite} \times \mathrm{adjacent} } \right) } = \frac { \left( \frac \mathrm{hypotenuse} \mathrm{adjacent} \right)} { \left( \frac \mathrm{hypotenuse} \mathrm{opposite} \right) } = \frac {\sec \theta}{\csc \theta}.$

$\cot \theta = \frac {\csc \theta}{\sec \theta}.$

[edit] 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:

$\sin\left( \pi/2-\theta\right) = \cos \theta$

$\cos\left( \pi/2-\theta\right) = \sin \theta$

$\tan\left( \pi/2-\theta\right) = \cot \theta$

$\cot\left( \pi/2-\theta\right) = \tan \theta$

$\sec\left( \pi/2-\theta\right) = \csc \theta$

$\csc\left( \pi/2-\theta\right) = \sec \theta$

[edit] 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:

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

$\tan^2 x + 1 = \sec^2 x \!\$

$\cot^2 x + 1 = \csc^2 x \!\$

Proof 3:

Differentiating the left-hand side of the identity yields:

2sin x cos x - 2sin 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.

[edit] Angle sum identities

[edit] Sine

Illustration of the sum formula.

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 = π/2 − RQP = π/2 − (π/2 − RQO) = RQO = α)

$OP = 1\,$

$PQ = \sin \beta\,$

$OQ = \cos \beta\,$

$\frac{AQ}{OQ} = \sin \alpha\,$ , so $AQ = \sin \alpha \cos \beta\,$

$\frac{PR}{PQ} = \cos \alpha\,$ , so $PR = \cos \alpha \sin \beta\,$

$\sin (\alpha + \beta) = PB = RB+PR = AQ+PR = \sin \alpha \cos \beta + \cos \alpha \sin \beta\,$

From the negative argument formulae, we also get

$\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta\,$

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

$e^{i\varphi}=\cos \varphi +i \cdot \sin \varphi$

So for angles : $α$ and : $β$ we have:

$e^{i\cdot (\alpha + \beta)} = \cos (\alpha +\beta) + i\cdot \sin(\alpha +\beta)$

Also using the following properties of exponential functions:

$e^{i(\alpha + \beta)} = e^{i \alpha}\cdot e^{i\beta}= (\cos \alpha +i\cdot \sin \alpha)\cdot (\cos \beta + i\cdot \sin \beta)$

Evaluating the product:

$e^{i(\alpha + \beta)} = (\cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta)+i\cdot (\sin \alpha \cdot \cos \beta + \sin \beta \cdot \cos \alpha)$

This will only be equal to the previous expression we got, if the imaginary and real parts are equal respectively. Hence we get:

$\cos (\alpha +\beta)=\cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta$

$\sin (\alpha +\beta)=\sin \alpha \cdot \cos \beta + \sin \beta \cdot \cos \alpha$

[edit] Cosine

Using the figure above,

$OP = 1\,$

$PQ = \sin \beta\,$

$OQ = \cos \beta\,$

$\frac{OA}{OQ} = \cos \alpha\,$ , so $OA = \cos \alpha \cos \beta\,$

$\frac{RQ}{PQ} = \sin \alpha\,$ , so $RQ = \sin \alpha \sin \beta\,$

$\cos (\alpha + \beta) = OB = OA-BA = OA-RQ = \cos \alpha \cos \beta\ - \sin \alpha \sin \beta\,$

From the negative argument formulae, we also get

$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\,$

Also, using the complementary angle formulae,

$\cos (\alpha + \beta) = \sin\left( \pi/2-(\alpha + \beta)\right) = \sin\left( (\pi/2-\alpha) - \beta\right)\,$

$= \sin\left( \pi/2-\alpha\right) \cos \beta - \cos\left( \pi/2-\alpha\right) \sin \beta\,$

$= \cos \alpha \cos \beta - \sin \alpha \sin \beta\,$

From the negative argument formulae, we also get

$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\,$

[edit] Tangent and cotangent

From the sine and cosine formulae, we get

$\tan (\alpha + \beta) = \frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)}\,$

$= \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta}\,$

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

$\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}\,$

$\tan (\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}\,$

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

$\cot (\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}\,$

$\cot (\alpha - \beta) = \frac{\cot \alpha \cot \beta + 1}{\cot \beta - \cot \alpha}\,$

[edit] Double-angle identities

From the angle sum identities, we get

$\sin (2 \theta) = 2 \sin \theta \cos \theta\,$

and

$\cos (2 \theta) = \cos^2 \theta - \sin^2 \theta\,$

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

$\cos (2 \theta) = 2 \cos^2 \theta - 1\,$

$\cos (2 \theta) = 1 - 2 \sin^2 \theta\,$

The angle sum identities also give

$\tan (2 \theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{2}{\cot \theta - \tan \theta}\,$

$\cot (2 \theta) = \frac{\cot^2 \theta - 1}{2 \cot \theta} = \frac{\cot \theta - \tan \theta}{2}\,$

[edit] Half-angle identities

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

$\cos \frac{\theta}{2} = \pm\, \sqrt\frac{1 + \cos \theta}{2},\,$

$\sin \frac{\theta}{2} = \pm\, \sqrt\frac{1 - \cos \theta}{2}.\,$

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

$\tan \frac{\theta}{2} = \pm\, \sqrt\frac{1 - \cos \theta}{1 + \cos \theta}.\,$

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

$\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}.\,$

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

$\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}.\,$

This also gives

$\tan \frac{\theta}{2} = \csc \theta - \cot \theta.\,$

Similar manipulations for the cot function give

$\cot \frac{\theta}{2} = \pm\, \sqrt\frac{1 + \cos \theta}{1 - \cos \theta} = \frac{1 + \cos \theta}{\sin \theta} = \frac{\sin \theta}{1 - \cos \theta} = \csc \theta + \cot \theta.\,$

[edit] Prosthaphaeresis identities

$\sin \theta \pm \sin \gamma = 2 \sin \frac{\theta\pm \gamma}2 \cos \frac{\theta\mp \gamma}2$
$\cos \theta + \cos \gamma = 2 \cos \frac{\theta+\gamma}2 \cos \frac{\theta-\gamma}2$
$\cos \theta - \cos \gamma = -2 \sin \frac{\theta+\gamma}2 \sin \frac{\theta-\gamma}2$

[edit] Inequalities

Illustration of the sine and tangent 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.

$OA = OD = 1\,$

$AB = \sin \theta\,$

$CD = \tan \theta\,$

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

$\sin \theta < \theta < \tan \theta\,$

This geometric argument applies if 0<θ<π/2. 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

$\frac{\sin \theta}{\theta} < 1\ \ \ \mathrm{if}\ \ \ 0 < \theta\,$

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

$\frac{\sin \theta}{\theta} = \frac{\sin (-\theta)}{-\theta} < 1\,$

Hence

$\frac{\sin \theta}{\theta} < 1\ \ \ \mathrm{if}\ \ \ \theta \ne 0\,$

$\frac{\tan \theta}{\theta} > 1\ \ \ \mathrm{if}\ \ \ 0 < \theta < \frac{\pi}{2}\,$

[edit] Identities involving calculus

[edit] Preliminaries

$\lim_{\theta \to 0}{\sin \theta} = 0\,$

$\lim_{\theta \to 0}{\cos \theta} = 1\,$

These can be seen from looking at the diagrams.

[edit] Sine and angle ratio identity

$\lim_{\theta \to 0}{\frac{\sin \theta}{\theta}} = 1$

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

$\sin \theta < \theta < \tan \theta\,$ , so

$\frac{\sin \theta}{\theta} < 1 < \frac{\tan \theta}{\theta}\,$ , so

$\frac{\sin \theta}{\theta \cos \theta} > 1\,$ , or

$\frac{\sin \theta}{\theta} > \cos \theta\,$ , so

$\cos \theta < \frac{\sin \theta}{\theta} < 1\,$ , but

$\lim_{\theta \to 0}{\cos \theta} = 1\,$ , so

$\lim_{\theta \to 0}{\frac{\sin \theta}{\theta}} = 1$

[edit] Cosine and angle ratio identity

$\lim_{\theta \to 0}\frac{1 - \cos \theta}{\theta} = 0\,$

Proof:

$\frac{1 - \cos \theta}{\theta} = \frac{1 - \cos^2 \theta}{\theta (1 + \cos \theta)}\,$

$= \frac{\sin^2 \theta}{\theta (1 + \cos \theta)}\,$

$= \frac{\sin \theta}{\theta} \times \sin \theta \times \frac{1}{1 + \cos \theta}\,$

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

[edit] Cosine and square of angle ratio identity

$\lim_{\theta \to 0}\frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$

Proof:

As in the preceding proof,

$\frac{1 - \cos \theta}{\theta^2} = \frac{\sin \theta}{\theta} \times \frac{\sin \theta}{\theta} \times \frac{1}{1 + \cos \theta}\,$

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

[edit] 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 sinθ < θ < tanθ 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—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.

[edit] References

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

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Proofs of trigonometric identities

From Wikipedia, the free encyclopedia

Contents

[edit] Elementary trigonometric identities

[edit] Definitions

[edit] Ratio identities

[edit] Complementary angle identities

[edit] Pythagorean identities

[edit] Angle sum identities

[edit] Sine

[edit] Cosine

[edit] Tangent and cotangent

[edit] Double-angle identities

[edit] Half-angle identities

[edit] Prosthaphaeresis identities

[edit] Inequalities

[edit] Identities involving calculus

[edit] Preliminaries

[edit] Sine and angle ratio identity

[edit] Cosine and angle ratio identity

[edit] Cosine and square of angle ratio identity

[edit] Pedagogical note

[edit] References

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