Difference between revisions of "Trigonometric identities"

Revision as of 11:39, 24 June 2006

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

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

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

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

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.

$\displaystyle \sin(\alpha + \beta) = \sin \alpha\cos \beta +\sin \beta \cos \alpha$	$\displaystyle \sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$
$\displaystyle \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$	$\displaystyle \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$
$\displaystyle \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$	$\displaystyle \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1+\tan \alpha \tan \beta}$

Double Angle Identities

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

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

Half Angle Identities

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

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

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

Law of Cosines

The Law of Cosines states

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

Law of Tangents

The Law of Tangents states

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

Other Identities

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

@@ Line 3: / Line 3: @@
 == Basic Definitions ==
 The six basic trigonometric functions can be defined using a right triangle:
+<center>[[Image:righttriangle.png]]</center>
-<center>[[Image:righttriangle.png]]</center>
 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 <math>\csc</math>).  They are defined as follows:
@@ Line 28: / Line 29: @@
 == 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> 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:
+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:
 {| style="height:150px; margin: 1em auto 1em auto"
@@ Line 44: / Line 45: @@
 {| style="width:100%; height:130px; margin: 1em auto 1em auto"
 |-
-| <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  \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> \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  \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> \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> \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>
 |}
@@ Line 56: / Line 57: @@
 {| style="height:200px; margin: 1em auto 1em auto"
 |-
-| <math> \sin 2\alpha </math> || = || <math>2\sin \alpha \cos \alpha</math>
+| <math> \displaystyle \sin 2\alpha </math> || = || <math> \displaystyle 2\sin \alpha \cos \alpha</math>
 |-
-| <math> \cos 2\alpha </math> || = || <math> \cos^2 \alpha - \sin^2 \alpha</math>
+| <math> \displaystyle \cos 2\alpha </math> || = || <math> \displaystyle \cos^2 \alpha - \sin^2 \alpha</math>
 |-
-| || = || <math> 2\cos^2 \alpha - 1</math>
+| || = || <math> \displaystyle 2\cos^2 \alpha - 1</math>
 |-
-| || = || <math> 1-2\sin^2 \alpha</math>
+| || = || <math> \displaystyle 1-2\sin^2 \alpha</math>
 |-
-| <math> \tan 2\alpha </math> || = || <math>\frac{2\tan \alpha}{1-\tan^2\alpha} </math>
+| <math> \displaystyle \tan 2\alpha </math> || = || <math>\frac{2\tan \alpha}{1-\tan^2\alpha} </math>
 |}
 == Half Angle Identities ==
-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.
+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.
 To summarize:
@@ Line 93: / Line 94: @@
 The extended [[Law of Sines]] states
-<center><math> \frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C} = 2R.</math></center>
+*<math> \displaystyle \frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C} = 2R.</math>
 == Law of Cosines ==
 The [[Law of Cosines]] states
-<center><math> a^2 = b^2 + c^2 - 2bc\cos A. </math></center>
+*<math> \displaystyle a^2 = b^2 + c^2 - 2bc\cos A. </math>
 == Law of Tangents ==
 The [[Law of Tangents]] states
-<center><math> \frac{b - c}{b + c} = \frac{\tan\frac 12(B-C)}{\tan \frac 12(B+C)}.</math></center>
+*<math> \displaystyle \frac{b - c}{b + c} = \frac{\tan\frac 12(B-C)}{\tan \frac 12(B+C)}.</math>
 == Other Identities ==

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