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Trigonometric identities

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Trigonometric identities are used to manipulate trigonometry equations in certain ways. Here is a list of them:

Contents

Basic Definitions

The six basic trigonometric functions can be defined using a right triangle:

Image:righttriangle.png

The six trig functions are sine, cosine, tangent, cosecant, secant, and cotangent. They are abbreviated by using the first three letters of their name (except for cosecant which uses ). They are defined as follows:

Even-Odd Identities

  • \sin (-\theta) = -\sin (\theta)
  • \tan (-\theta) = -\tan (\theta)
  • \csc (-\theta) = -\csc (\theta)
  • \cot (-\theta) = -\cot (\theta)

Reciprocal Relations

From the last section, it is easy to see that the following hold:

Another useful identity that isn't a reciprocal relation is that .

Note that ; the former refers to the inverse trigonometric functions.

Pythagorean Identities

Using the Pythagorean Theorem on our triangle above, we know that . If we divide by we get \left(\frac ac\right)^2 + \left(\frac bc\right)^2 = 1, which is just . Dividing by or instead produces two other similar identities. The Pythagorean Identities are listed below:

(Note that the second two are easily derived by dividing the first by and )

Angle Addition/Subtraction Identities

Once we have formulas for angle addition, angle subtraction is rather easy to derive. For example, we just look at and we can derive the sine angle subtraction formula using the sine angle addition formula.

  • \sin(\alpha + \beta) = \sin \alpha\cos \beta +\sin \beta \cos \alpha || \sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha
  • \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta || \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta
  • \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}

We can prove \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta easily by using \sin(\alpha + \beta) = \sin \alpha\cos \beta +\sin \beta \cos \alpha and .


= \sin (90- \alpha) \cos (\beta) - \sin ( \beta) \cos (90- \alpha)

=\cos \alpha \cos \beta - \sin \beta \sin \alpha

Double Angle Identities

Double angle identities are easily derived from the angle addition formulas by just letting . Doing so yields:

\begin{eqnarray*}\sin 2\alpha &=& 2\sin \alpha \cos \alpha\\\cos 2\alpha  &=& \cos^2 \alpha - \sin^2 \alpha\\&=& 2\cos^2 \alpha - 1\\&=& 1-2\sin^2 \alpha\\\tan 2\alpha  &=& \frac{2\tan \alpha}{1-\tan^2\alpha}

Half Angle Identities

Using the double angle identities, we can now derive half angle identities. The double angle formula for cosine tells us \cos 2\alpha = 2\cos^2 \alpha - 1. Solving for we get \cos \alpha =\pm \sqrt{\frac{1 + \cos 2\alpha}2} where we look at the quadrant of to decide if it's positive or negative. Likewise, we can use the fact that \cos 2\alpha = 1 - 2\sin^2 \alpha to find a half angle identity for sine. Then, to find a half angle identity for tangent, we just use the fact that \tan \frac x2 =\frac{\sin \frac x2}{\cos \frac x2} and plug in the half angle identities for sine and cosine.

To summarize:

  • \sin \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}2}
  • \cos \frac{\theta}2 = \pm \sqrt{\frac{1+\cos \theta}2}
  • \tan \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}

Prosthaphaeresis Identities

(Otherwise known as sum-to-product 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

Law of Sines

Main article: Law of Sines

The extended Law of Sines states

  • \frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C} = 2R.

Law of Cosines

Main article: Law of Cosines

The Law of Cosines states

Law of Tangents

Main article: Law of Tangents

The Law of Tangents states that if and are angles in a triangle opposite sides and respectively, then

\frac{\tan{\left(\frac{A-B}{2}\right)}}{\tan{\left(\frac{A+B}{2}\right)}}=\frac{a-b}{a+b} .

Other Identities

  • e^{i\theta} = \cos \theta + i\sin \theta (This is also written as )
  • |1-e^{i\theta}|=2\sin\frac{\theta}{2}

See also

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