Difference between revisions of "Trigonometric identities"

Revision as of 15:49, 7 October 2007

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

Basic Definitions

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

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 $\csc$ ). They are defined as follows:

$\sin A = \frac ac$	$\csc A = \frac ca$
$\cos A = \frac bc$	$\sec A = \frac cb$
$\tan A = \frac ab$	$\cot A = \frac ba$

Reciprocal Relations

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

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

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

$\tan A = \frac 1{\cot A}$

Another useful identity that isn't a reciprocal relation is that $\tan A =\frac{\sin A}{\cos A}$ .

Pythagorean Identities

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

$\sin^2x + \cos^2x = 1$

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

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

(Note that the second two are easily derived by dividing the first by $\cos^2x$ and $\sin^2x$ )

Angle Addition/Subtraction Identities

Once we have formulas for angle addition, angle subtraction is rather easy to derive. For example, we just look at $\sin(\alpha+(-\beta))$ 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 $\sin(x)=\cos(90-x)$ $\cos (\alpha + \beta) = \sin((90 -\alpha) - \beta) <cmath>= \sin (90- \alpha) \cos (\beta) - \sin ( \beta) \cos (90- \alpha) </cmath>=\cos \alpha \cos \beta - \sin \beta \sin \alpha$

Double Angle Identities

Double angle identities are easily derived from the angle addition formulas by just letting $\alpha = \beta$ . Doing so yields:

$\sin 2\alpha$	=	$2\sin \alpha \cos \alpha$
$\cos 2\alpha$	=	$\cos^2 \alpha - \sin^2 \alpha$
	=	$2\cos^2 \alpha - 1$
	=	$1-2\sin^2 \alpha$
$\tan 2\alpha$	=	$\frac{2\tan \alpha}{1-\tan^2\alpha}$

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 $\cos \alpha$ we get $\cos \alpha =\pm \sqrt{\frac{1 + \cos 2\alpha}2}$ where we look at the quadrant of $\alpha$ 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:

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

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

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

Even-Odd Identities

$\sin (-\theta) = -\sin (\theta)$

$\cos (-\theta) = \cos (\theta)$

$\tan (-\theta) = -\tan (\theta)$

$\csc (-\theta) = -\csc (\theta)$

$\sec (-\theta) = \sec (\theta)$

$\cot (-\theta) = -\cot (\theta)$

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

The extended Law of Sines states

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

Law of Cosines

The Law of Cosines states

$a^2 = b^2 + c^2 - 2bc\cos A.$

Law of Tangents

The Law of Tangents states

$\frac{b - c}{b + c} = \frac{\tan\frac 12(B-C)}{\tan \frac 12(B+C)}.$

Other Identities

$|1-e^{i\theta}|=2\sin\frac{\theta}{2}$

@@ Line 33: / Line 33: @@
 == Pythagorean Identities ==
-Using the [[Pythagorean Theorem]] on our triangle above, we know that <math>\displaystyle a^2 + b^2 = c^2 </math>.  If we divide by <math> \displaystyle c^2 </math> we get <math> \displaystyle \left(\frac ac\right)^2 + \left(\frac bc\right)^2 = 1 </math> which is just <math> \displaystyle \sin^2 A + \cos^2 A =1 </math>.  Dividing by <math>\displaystyle  a^2 </math> or <math>\displaystyle  b^2 </math> instead produces two other similar identities.  The Pythagorean Identities are listed below:
+Using the [[Pythagorean Theorem]] on our triangle above, we know that <math>a^2 + b^2 = c^2 </math>.  If we divide by <math>c^2 </math> we get <math>\left(\frac ac\right)^2 + \left(\frac bc\right)^2 = 1 </math> which is just <math>\sin^2 A + \cos^2 A =1 </math>.  Dividing by <math> a^2 </math> or <math> b^2 </math> instead produces two other similar identities.  The Pythagorean Identities are listed below:
 {| style="height:150px; margin: 1em auto 1em auto"
 |-
-|<math>\displaystyle \sin^2x + \cos^2x = 1</math>
+|<math>\sin^2x + \cos^2x = 1</math>
 |-
-|<math>\displaystyle 1 + \cot^2x = \csc^2x</math>
+|<math>1 + \cot^2x = \csc^2x</math>
 |-
-|<math>\displaystyle \tan^2x + 1 = \sec^2x</math>
+|<math>\tan^2x + 1 = \sec^2x</math>
 |}
@@ Line 51: / Line 51: @@
 {| style="width:100%; height:130px; margin: 1em auto 1em auto"
 |-
-| <math> \displaystyle  \sin(\alpha + \beta) = \sin \alpha\cos \beta +\sin \beta \cos \alpha</math> || <math> \displaystyle \sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha</math>
+| <math> \sin(\alpha + \beta) = \sin \alpha\cos \beta +\sin \beta \cos \alpha</math> || <math>\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha</math>
 |-
-| <math> \displaystyle  \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta </math> || <math> \displaystyle \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta</math>
+| <math> \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta </math> || <math>\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta</math>
 |-
-| <math> \displaystyle \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta} </math> || <math> \displaystyle \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1+\tan \alpha \tan \beta} </math>
+| <math>\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta} </math> || <math>\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1+\tan \alpha \tan \beta} </math>
 |}
-We can prove <math> \displaystyle  \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta </math> easily by using <math> \displaystyle  \sin(\alpha + \beta) = \sin \alpha\cos \beta +\sin \beta \cos \alpha</math> and <math>\sin(x)=\cos(90-x)</math>
+We can prove <math> \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta </math> easily by using <math> \sin(\alpha + \beta) = \sin \alpha\cos \beta +\sin \beta \cos \alpha</math> and <math>\sin(x)=\cos(90-x)</math>
-<math>\displaystyle \cos (\alpha + \beta) = \sin((90 -\alpha) - \beta) </math><math>\displaystyle = \sin (90- \alpha) \cos (\beta) - \sin ( \beta) \cos (90- \alpha) </math><math>=\displaystyle \cos \alpha \cos \beta - \sin \beta \sin \alpha </math>
+<math>\cos (\alpha + \beta) = \sin((90 -\alpha) - \beta) <cmath>= \sin (90- \alpha) \cos (\beta) - \sin ( \beta) \cos (90- \alpha) </cmath>=\cos \alpha \cos \beta - \sin \beta \sin \alpha </math>
 == Double Angle Identities ==
@@ Line 65: / Line 65: @@
 {| style="height:200px; margin: 1em auto 1em auto"
 |-
-| <math> \displaystyle \sin 2\alpha </math> || = || <math> \displaystyle 2\sin \alpha \cos \alpha</math>
+| <math>\sin 2\alpha </math> || = || <math>2\sin \alpha \cos \alpha</math>
 |-
-| <math> \displaystyle \cos 2\alpha </math> || = || <math> \displaystyle \cos^2 \alpha - \sin^2 \alpha</math>
+| <math>\cos 2\alpha </math> || = || <math>\cos^2 \alpha - \sin^2 \alpha</math>
 |-
-| || = || <math> \displaystyle 2\cos^2 \alpha - 1</math>
+| || = || <math>2\cos^2 \alpha - 1</math>
 |-
-| || = || <math> \displaystyle 1-2\sin^2 \alpha</math>
+| || = || <math>1-2\sin^2 \alpha</math>
 |-
-| <math> \displaystyle \tan 2\alpha </math> || = || <math>\frac{2\tan \alpha}{1-\tan^2\alpha} </math>
+| <math>\tan 2\alpha </math> || = || <math>\frac{2\tan \alpha}{1-\tan^2\alpha} </math>
 |}
@@ Line 79: / Line 79: @@
 == Half Angle Identities ==
-Using the double angle identities, we can now derive half angle identities.  The double angle formula for cosine tells us <math> \displaystyle \cos 2\alpha = 2\cos^2 \alpha - 1 </math>.  Solving for <math> \displaystyle \cos \alpha </math> we get <math> \displaystyle \cos \alpha =\pm \sqrt{\frac{1 + \cos 2\alpha}2}</math> where we look at the quadrant of <math> \displaystyle \alpha </math> to decide if it's positive or negative.  Likewise, we can use the fact that <math> \displaystyle \cos 2\alpha = 1 - 2\sin^2 \alpha </math> to find a half angle identity for sine.  Then, to find a half angle identity for tangent, we just use the fact that <math> \displaystyle \tan \frac x2 =\frac{\sin \frac x2}{\cos \frac x2} </math> and plug in the half angle identities for sine and cosine.
+Using the double angle identities, we can now derive half angle identities.  The double angle formula for cosine tells us <math>\cos 2\alpha = 2\cos^2 \alpha - 1 </math>.  Solving for <math>\cos \alpha </math> we get <math>\cos \alpha =\pm \sqrt{\frac{1 + \cos 2\alpha}2}</math> where we look at the quadrant of <math>\alpha </math> to decide if it's positive or negative.  Likewise, we can use the fact that <math>\cos 2\alpha = 1 - 2\sin^2 \alpha </math> to find a half angle identity for sine.  Then, to find a half angle identity for tangent, we just use the fact that <math>\tan \frac x2 =\frac{\sin \frac x2}{\cos \frac x2} </math> and plug in the half angle identities for sine and cosine.
 To summarize:
@@ Line 95: / Line 95: @@
 == Even-Odd Identities ==
-<math> \displaystyle \sin (-\theta) = -\sin (\theta) </math>
+<math>\sin (-\theta) = -\sin (\theta) </math>
-<math> \displaystyle \cos (-\theta) = \cos (\theta) </math>
+<math>\cos (-\theta) = \cos (\theta) </math>
-<math> \displaystyle \tan (-\theta) = -\tan (\theta) </math>
+<math>\tan (-\theta) = -\tan (\theta) </math>
-<math> \displaystyle \csc (-\theta) = -\csc (\theta) </math>
+<math>\csc (-\theta) = -\csc (\theta) </math>
-<math> \displaystyle \sec (-\theta) = \sec (\theta) </math>
+<math>\sec (-\theta) = \sec (\theta) </math>
-<math> \displaystyle \cot (-\theta) = -\cot (\theta) </math>
+<math>\cot (-\theta) = -\cot (\theta) </math>
@@ Line 121: / Line 121: @@
 The extended [[Law of Sines]] states
-*<math> \displaystyle \frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C} = 2R.</math>
+*<math>\frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C} = 2R.</math>
@@ Line 128: / Line 128: @@
 The [[Law of Cosines]] states
-*<math> \displaystyle a^2 = b^2 + c^2 - 2bc\cos A. </math>
+*<math>a^2 = b^2 + c^2 - 2bc\cos A. </math>
@@ Line 135: / Line 135: @@
 The [[Law of Tangents]] states
-*<math> \displaystyle \frac{b - c}{b + c} = \frac{\tan\frac 12(B-C)}{\tan \frac 12(B+C)}.</math>
+*<math>\frac{b - c}{b + c} = \frac{\tan\frac 12(B-C)}{\tan \frac 12(B+C)}.</math>
@@ Line 149: / Line 149: @@
 * [[Trigonometric substitution]]
 * [http://www.sosmath.com/trig/Trig5/trig5/trig5.html Trigonometric Identities]
+[[Category:Trigonometry]]

Page

Toolbox

Search