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Difference between revisions of "Trigonometric identities"

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'''Trigonometric identities''' are used to manipulate [[trigonometry]] [[equation]]s in certain ways.  Here is a list of them:
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'''Trigonometric Identities''' are used to manipulate [[trigonometry]] [[equation]]s in certain ways.  Here is a list of them:
  
 
== Basic Definitions ==
 
== Basic Definitions ==
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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:
 
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:
−
 
+
{| class="wikitable"
−
*<math>\sin A = \frac ac</math>   
+
|+ Basic Definitions
−
*<math>\csc A = \frac ca</math>
+
|- <!-- Start of a new row -->
−
*<math> \cos A = \frac bc</math>  
+
| <math>\sin A = \frac ac</math>  || <math>\csc A = \frac ca</math> || <math> \cos A = \frac bc</math> || <math>\sec A = \frac cb</math> || <math> \tan A = \frac ab</math> || <math> \cot A = \frac ba</math>
−
*<math>\sec A = \frac cb</math>
+
|}
−
*<math> \tan A = \frac ab</math>  
 
−
*<math> \cot A = \frac ba</math>
 
  
 
== Even-Odd Identities ==
 
== Even-Odd Identities ==
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*<math>\tan (-\theta) = -\tan (\theta) </math>
 
*<math>\tan (-\theta) = -\tan (\theta) </math>
 +
 +
*<math>\sec (-\theta) = \sec (\theta) </math>
  
 
*<math>\csc (-\theta) = -\csc (\theta) </math>
 
*<math>\csc (-\theta) = -\csc (\theta) </math>
  
−
*<math>\sec (-\theta) = \sec (\theta) </math>
+
*<math>\cot (-\theta) = -\cot (\theta) </math>
 +
 
 +
===Further Conclusions===
 +
 
 +
Based on the above identities, we can also claim that
 +
 
 +
*<math>\sin(\cos(-\theta)) = \sin(\cos(\theta))</math>
 +
 
 +
*<math>\cos(\sin(-\theta)) = \cos(-\sin(\theta)) = \cos(\sin(\theta))</math>
  
−
*<math>\cot (-\theta) = -\cot (\theta) </math>
+
This is only true when <math>\sin(\theta)</math> is in the domain of <math>\cos(\theta)</math>.
  
 
== Reciprocal Relations ==
 
== Reciprocal Relations ==
−
From the last section, it is easy to see that the following hold:
+
From the first section, it is easy to see that the following hold:
  
 
*<math> \sin A = \frac 1{\csc A}</math>  
 
*<math> \sin A = \frac 1{\csc A}</math>  
 +
 
*<math> \cos A = \frac 1{\sec A}</math>
 
*<math> \cos A = \frac 1{\sec A}</math>
 +
 
*<math> \tan A = \frac 1{\cot A}</math>
 
*<math> \tan A = \frac 1{\cot A}</math>
  
 
Another useful identity that isn't a reciprocal relation is that <math> \tan A =\frac{\sin A}{\cos A} </math>.
 
Another useful identity that isn't a reciprocal relation is that <math> \tan A =\frac{\sin A}{\cos A} </math>.
  
−
Note that <math>\sin^{-1} A \neq \csc A</math>; the former refers to the [[inverse trigonometric function]]s.  
+
Note that <math>\sin^{-1} A \neq \csc A</math>; the former refers to the [[inverse trigonometric function]]s.
  
 
== Pythagorean Identities ==
 
== Pythagorean Identities ==
−
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:
+
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{a}{c}\right)^2 + \left(\frac{b}{c}\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:
  
 
*<math>\sin^2x + \cos^2x = 1</math>
 
*<math>\sin^2x + \cos^2x = 1</math>
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*<math>\tan^2x + 1 = \sec^2x</math>
 
*<math>\tan^2x + 1 = \sec^2x</math>
  
−
(Note that the second two are easily derived by dividing the first by <math>\cos^2x</math> and <math>\sin^2x</math>)
+
(Note that the last two are easily derived by dividing the first by <math>\sin^2x</math> and <math>\cos^2x</math>, respectively.)
  
 
== Angle Addition/Subtraction Identities ==
 
== Angle Addition/Subtraction Identities ==
 
Once we have formulas for angle addition, angle subtraction is rather easy to derive.  For example, we just look at <math> \sin(\alpha+(-\beta))</math> and we can derive the sine angle subtraction formula using the sine angle addition formula.
 
Once we have formulas for angle addition, angle subtraction is rather easy to derive.  For example, we just look at <math> \sin(\alpha+(-\beta))</math> and we can derive the sine angle subtraction formula using the sine angle addition formula.
  
−
*<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> \sin(\alpha \pm \beta) = \sin \alpha\cos \beta \pm\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> \cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \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>\tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1\mp\tan \alpha \tan \beta} </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>.
 
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>\cos (\alpha + \beta)</math>
 
<math>\cos (\alpha + \beta)</math>
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Double angle identities are easily derived from the angle addition formulas by just letting <math> \alpha = \beta </math>.  Doing so yields:
 
Double angle identities are easily derived from the angle addition formulas by just letting <math> \alpha = \beta </math>.  Doing so yields:
  
−
<math>\begin{eqnarray*}
+
<cmath>\begin{eqnarray*}
 
\sin 2\alpha &=& 2\sin \alpha \cos \alpha\\
 
\sin 2\alpha &=& 2\sin \alpha \cos \alpha\\
 
\cos 2\alpha  &=& \cos^2 \alpha - \sin^2 \alpha\\
 
\cos 2\alpha  &=& \cos^2 \alpha - \sin^2 \alpha\\
 
&=& 2\cos^2 \alpha - 1\\
 
&=& 2\cos^2 \alpha - 1\\
 
&=& 1-2\sin^2 \alpha\\
 
&=& 1-2\sin^2 \alpha\\
−
\tan 2\alpha  &=& \frac{2\tan \alpha}{1-\tan^2\alpha} </math>
+
\tan 2\alpha  &=& \frac{2\tan \alpha}{1-\tan^2\alpha} \end{eqnarray*}</cmath>
 +
 
 +
=Further Conclusions=
 +
 
 +
We can see from the above that
 +
 
 +
*<math>\csc(2a) = \frac{\csc(a)\sec(a)}{2}</math>
 +
*<math>\sec(2a) = \frac{1}{2\cos^2(a) - 1} = \frac{1}{\cos^2(a) - \sin^2(a)} = \frac{1}{1 - 2\sin^2(a)}</math>
 +
*<math>\cot(2a) = \frac{1 - \tan^2(a)}{2\tan(a)}</math>
  
 
== Half Angle Identities ==
 
== 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 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:
 
To summarize:
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*<math> \sin \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}2} </math>
 
*<math> \sin \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}2} </math>
 
*<math> \cos \frac{\theta}2 = \pm \sqrt{\frac{1+\cos \theta}2} </math>
 
*<math> \cos \frac{\theta}2 = \pm \sqrt{\frac{1+\cos \theta}2} </math>
−
*<math> \tan \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} </math>
+
*<math> \tan \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\frac{\sin \theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin \theta}  </math>
  
 
== Prosthaphaeresis Identities ==
 
== Prosthaphaeresis Identities ==
 
(Otherwise known as sum-to-product identities)
 
(Otherwise known as sum-to-product identities)
  
−
* <math>\sin \theta \pm \sin \gamma = 2 \sin \frac{\theta\pm \gamma}2 \cos \frac{\theta\mp \gamma}2</math>
+
* <math>{\sin \theta + \sin \gamma = 2 \sin \frac{\theta + \gamma}2 \cos \frac{\theta - \gamma}2}</math>
−
* <math>\cos \theta + \cos \gamma = 2 \cos \frac{\theta+\gamma}2 \cos \frac{\theta-\gamma}2</math>
+
* <math>{\sin \theta - \sin \gamma = 2 \sin \frac{\theta - \gamma}2 \cos \frac{\theta + \gamma}2}</math>
−
* <math>\cos \theta - \cos \gamma = -2 \sin \frac{\theta+\gamma}2 \sin \frac{\theta-\gamma}2</math>
+
* <math>{\cos \theta + \cos \gamma = 2 \cos \frac{\theta+\gamma}2 \cos \frac{\theta-\gamma}2}</math>
 +
* <math>{\cos \theta - \cos \gamma = -2 \sin \frac{\theta+\gamma}2 \sin \frac{\theta-\gamma}2}</math>
  
 
== Law of Sines ==
 
== Law of Sines ==
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The extended [[Law of Sines]] states
 
The extended [[Law of Sines]] states
  
−
*<math>\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>
  
 
== Law of Cosines ==
 
== Law of Cosines ==
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== Law of Tangents ==
 
== Law of Tangents ==
 
{{main|Law of Tangents}}
 
{{main|Law of Tangents}}
−
The [[Law of Tangents]] states
+
The [[Law of Tangents]] states that if <math>A</math> and <math>B</math> are angles in a triangle opposite sides <math>a</math> and <math>b</math> respectively, then
  
−
*<math>\frac{b - c}{b + c} = \frac{\tan\frac 12(B-C)}{\tan \frac 12(B+C)}.</math>
+
<math> \frac{\tan{\left(\frac{A-B}{2}\right)}}{\tan{\left(\frac{A+B}{2}\right)}}=\frac{a-b}{a+b} . </math>
 +
 
 +
A further extension of the [[Law of Tangents]] states that if <math>A</math>, <math>B</math>, and <math>C</math> are angles in a triangle, then
 +
<math>\tan(A)\cdot\tan(B)\cdot\tan(C)=\tan(A)+\tan(B)+\tan(C)</math>
  
 
== Other Identities ==
 
== Other Identities ==
−
*<math>e^{i\theta} = \cos \theta + i\sin \theta</math>
+
*<math>\sin(90-\theta) = \cos(\theta)</math>
 +
*<math>\cos(90-\theta)=\sin(\theta)</math>
 +
*<math>\tan(90-\theta)=\cot(\theta)</math>
 +
*<math>\sin(180-\theta) = \sin(\theta)</math>
 +
*<math>\cos(180-\theta) = -\cos(\theta)</math>
 +
*<math>\tan(180-\theta) = -\tan(\theta)</math>
 +
*<math>e^{i\theta} = \cos \theta + i\sin \theta</math> (This is also written as <math>\text{cis }\theta</math>)
 
*<math>|1-e^{i\theta}|=2\sin\frac{\theta}{2}</math>
 
*<math>|1-e^{i\theta}|=2\sin\frac{\theta}{2}</math>
 +
*<math>\left(\tan\theta + \sec\theta\right)^2 = \frac{1 + \sin\theta}{1 - \sin\theta}</math>
 +
*<math>\sin(\theta) = \cos(\theta)\tan(\theta)</math>
 +
*<math>\cos(\theta) = \frac{\sin(\theta)}{\tan(\theta)}</math>
 +
*<math>\sec(\theta) = \frac{\tan(\theta)}{\sin(\theta)}</math>
 +
*<math>\arctan(x) + \arctan(y) = \arctan \left( \dfrac{x+y}{1-xy} \right)</math>
 +
*<math>\sin^2(\theta) + \cos^2(\theta) + \tan^2(\theta) = \sec^2(\theta)</math>
 +
*<math>\sin^2(\theta) + \cos^2(\theta) + \cot^2(\theta) = \csc^2(\theta)</math>
 +
 +
The two identities above are derived from the Pythagorean Identities.
 +
 +
*<math>\cos(2\theta) = (\cos(\theta) + \sin(\theta))(\cos(\theta) - \sin(\theta))</math>
  
 
==See also==
 
==See also==
 
* [[Trigonometry]]
 
* [[Trigonometry]]
 
* [[Trigonometric substitution]]
 
* [[Trigonometric substitution]]
−
* [http://www.sosmath.com/trig/Trig5/trig5/trig5.html Trigonometric Identities]
+
 
 +
==External Links==
 +
[http://www.sosmath.com/trig/Trig5/trig5/trig5.html Trigonometric Identities]
  
 
[[Category:Trigonometry]]
 
[[Category:Trigonometry]]

Revision as of 11:03, 30 July 2020

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

Basic Definitions

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

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

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

Even-Odd Identities

  • $\sin (-\theta) = -\sin (\theta)$
  • $\cos (-\theta) = \cos (\theta)$
  • $\tan (-\theta) = -\tan (\theta)$
  • $\sec (-\theta) = \sec (\theta)$
  • $\csc (-\theta) = -\csc (\theta)$
  • $\cot (-\theta) = -\cot (\theta)$

Further Conclusions

Based on the above identities, we can also claim that

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

This is only true when $\sin(\theta)$ is in the domain of $\cos(\theta)$.

Reciprocal Relations

From the first 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 $\tan A =\frac{\sin A}{\cos A}$.

Note that $\sin^{-1} A \neq \csc A$; the former refers to the inverse trigonometric functions.

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 $\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1$, 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 last two are easily derived by dividing the first by $\sin^2x$ and $\cos^2x$, respectively.)

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.

  • $\sin(\alpha \pm \beta) = \sin \alpha\cos \beta \pm\sin \beta \cos \alpha$
  • $\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$
  • $\tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1\mp\tan \alpha \tan \beta}$

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)$$= \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 $\alpha = \beta$. 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} \end{eqnarray*}

Further Conclusions

We can see from the above that

  • $\csc(2a) = \frac{\csc(a)\sec(a)}{2}$
  • $\sec(2a) = \frac{1}{2\cos^2(a) - 1} = \frac{1}{\cos^2(a) - \sin^2(a)} = \frac{1}{1 - 2\sin^2(a)}$
  • $\cot(2a) = \frac{1 - \tan^2(a)}{2\tan(a)}$

Half Angle Identities

Using the double angle identities, we can 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:

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

Prosthaphaeresis Identities

(Otherwise known as sum-to-product identities)

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

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

  • $a^2 = b^2 + c^2 - 2bc\cos A.$

Law of Tangents

Main article: Law of Tangents

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

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

A further extension of the Law of Tangents states that if $A$, $B$, and $C$ are angles in a triangle, then $\tan(A)\cdot\tan(B)\cdot\tan(C)=\tan(A)+\tan(B)+\tan(C)$

Other Identities

  • $\sin(90-\theta) = \cos(\theta)$
  • $\cos(90-\theta)=\sin(\theta)$
  • $\tan(90-\theta)=\cot(\theta)$
  • $\sin(180-\theta) = \sin(\theta)$
  • $\cos(180-\theta) = -\cos(\theta)$
  • $\tan(180-\theta) = -\tan(\theta)$
  • $e^{i\theta} = \cos \theta + i\sin \theta$ (This is also written as $\text{cis }\theta$)
  • $|1-e^{i\theta}|=2\sin\frac{\theta}{2}$
  • $\left(\tan\theta + \sec\theta\right)^2 = \frac{1 + \sin\theta}{1 - \sin\theta}$
  • $\sin(\theta) = \cos(\theta)\tan(\theta)$
  • $\cos(\theta) = \frac{\sin(\theta)}{\tan(\theta)}$
  • $\sec(\theta) = \frac{\tan(\theta)}{\sin(\theta)}$
  • $\arctan(x) + \arctan(y) = \arctan \left( \dfrac{x+y}{1-xy} \right)$
  • $\sin^2(\theta) + \cos^2(\theta) + \tan^2(\theta) = \sec^2(\theta)$
  • $\sin^2(\theta) + \cos^2(\theta) + \cot^2(\theta) = \csc^2(\theta)$

The two identities above are derived from the Pythagorean Identities.

  • $\cos(2\theta) = (\cos(\theta) + \sin(\theta))(\cos(\theta) - \sin(\theta))$

See also

External Links

Trigonometric Identities

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Trigonometric_identities&oldid=129810"