Difference between revisions of "Trigonometric identities"

Revision as of 21:27, 29 May 2021

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

Even-Odd Identities

The functions $\sin(x)$ , $\tan(x)$ , and $\csc(x)$ are odd, while $cos(x)$ , $\cot(x)$ , and $\sec(x)$ are even. In other words, the six trigonometric functions satisfy the following equalities:

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

These are derived by the unit circle definitions of trigonometry. Making any angle negative is the same as flipping it across the x-axis. This keeps its x-coordinate the same, but makes its y-coordinate negative. Thus, $\sin(-x) = -\sin(x)$ and $\cos(-x) = \cos(x)$ .

Pythagorean Identities

The Pythagorean identities state that

$\sin^2x + \cos^2x = 1$
$1 + \cot^2x = \csc^2x$
$\tan^2x + 1 = \sec^2x$

Using the ratio definition of trigonometry, we apply Pythagorean Theorem on our triangle above to get 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 $\sin^2 x + \cos^2 x =1$ . Using the unit circle definition of trigonometry, it's extremely easy to see that $\sin^2 x+ \cos^2 x =1$ .

To derive the other two, divide by either $\sin^2 (x)$ or $\cos^2 (x)$ and substitute the respective trigonometry in place of the ratios.

Angle Addition Identities

The trigonometric angle addition identities state the following identities:

$\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)$
$\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)$
$\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)}$

There are many proofs of these identities. For the sake of brevity, we list only one here.

Euler's identity states that $e^{ix} = \cos (x) + i \sin(x)$ . We have that $\begin{align*} \cos (x+y) + i \sin (x+y) &= e^{i(x+y)} \\ &= e^{ix} \cdot e^{iy} \\ &= (\cos (x) + i \sin (x))(\cos (y) + i \sin (y)) \\ &= (\cos (x) \cos (y) - \sin (x) \sin(y)) + i(\sin (x) \cos(y) + \cos(x) \sin(y)) \end{align*}$ By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas.

To derive the tangent addition formula, we reduce tangent to sine and cosine, divide both numerator and denominator by $\cos (x) \cos (y)$ , and simplify. $\begin{align*} \tan (x+y) &= \frac{\sin (x+y)}{\cos (x+y)} \\ &= \frac{\sin (x) \cos(y) + \cos(x) \sin(y)}{\cos (x) \cos (y) - \sin (x) \sin(y)} \\ &= \frac{\frac{\sin(x)}{\cos(x)} + \frac{\sin(y)}{\cos(x)}}{1 - \frac{\sin (x) \sin(y)}{\cos (x) \cos(y)}} \\ &= \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan(y)} \end{align*}$ as desired.

Double-Angle Identities

The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting $x = y$ . Doing so yields:

$\sin (2x) = 2\sin (x) \cos (x)$
$\cos (2x) = \cos^2 (x) - \sin^2 (x)$
$\tan (2x) = \frac{2\tan (x)}{1-\tan^2 (x)}$

Cosine Double-Angle

Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle identity given above.

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

In addition, the following identities are useful in integration and in deriving the half-angle identities. They are a simple rearrangement of the two above.

$\sin^2 (x) = \frac{1 - \cos (2x)}{2}$
$\cos^2 (x) = \frac{1 + \cos (2x)}{2}$

Half-Angle Identities

The trigonometric half-angle identities state the following equalities:

$\sin (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{2}}$
$\cos (\frac{x}{2}) = \pm \sqrt{\frac{1 + \cos (x)}{2}}$
$\tan (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{1+\cos \theta}} = \frac{\sin x}{1 + \cos (x)} = \frac{1-\cos (x)}{\sin (X)}$

The plus or minus is not saying that there are two answers, but that the sine of the angle depends on the quadrant that the angle is in.

Take the two expressions listed in the cosine double-angle section for $\sin^2 (x)$ and $\cos^2 (x)$ , and substitute $\frac{1}{2} x$ instead of $x$ . Taking the square root then yields the desired half-angle identities for sine and cosine.

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.

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

External Links

Trigonometric Identities

@@ Line 1: / Line 1: @@
 '''Trigonometric Identities''' are used to manipulate [[trigonometry]] [[equation]]s in certain ways.  Here is a list of them:
-== Basic Definitions ==
-The six basic trigonometric functions can be defined using a right triangle:
-<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:
-{| class="wikitable"
-|+ Basic Definitions
-|- <!-- Start of a new row -->
-| <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>
-|}
 == Even-Odd Identities ==
-*<math>\sin (-\theta) = -\sin (\theta) </math>
+The functions <math>\sin(x)</math>, <math>\tan(x)</math>, and <math>\csc(x)</math> are odd, while <math>cos(x)</math>, <math>\cot(x)</math>, and <math>\sec(x)</math> are even. In other words, the six trigonometric functions satisfy the following equalities:
+* <math>\sin (-x) = -\sin (x) </math>
-*<math>\cos (-\theta) = \cos (\theta) </math>
+* <math>\cos (-x) = \cos (x) </math>
+* <math>\tan (-x) = -\tan (x) </math>
-*<math>\tan (-\theta) = -\tan (\theta) </math>
+* <math>\sec (-x) = \sec (x) </math>
+* <math>\csc (-x) = -\csc (x) </math>
-*<math>\sec (-\theta) = \sec (\theta) </math>
+* <math>\cot (-x) = -\cot (x) </math>
-*<math>\csc (-\theta) = -\csc (\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>
+These are derived by the unit circle definitions of trigonometry. Making any angle negative is the same as flipping it across the x-axis. This keeps its x-coordinate the same, but makes its y-coordinate negative. Thus, <math>\sin(-x) = -\sin(x)</math> and <math>\cos(-x) = \cos(x)</math>.
-This is only true when <math>\sin(\theta)</math> is in the domain of <math>\cos(\theta)</math>.
-== Reciprocal Relations ==
-From the first section, it is easy to see that the following hold:
-*<math> \sin A = \frac 1{\csc A}</math>
-*<math> \cos A = \frac 1{\sec 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>.
-Note that <math>\sin^{-1} A \neq \csc A</math>; the former refers to the [[inverse trigonometric function]]s.
 == 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{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:
+The Pythagorean identities state that
+* <math>\sin^2x + \cos^2x = 1</math>
-*<math>\sin^2x + \cos^2x = 1</math>
+* <math>1 + \cot^2x = \csc^2x</math>
-*<math>1 + \cot^2x = \csc^2x</math>
+* <math>\tan^2x + 1 = \sec^2x</math>
-*<math>\tan^2x + 1 = \sec^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 ==
-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 \pm \beta) = \sin \alpha\cos \beta \pm\sin \beta \cos \alpha</math>
-*<math> \cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \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>.
-<math>\cos (\alpha + \beta)</math>
-<math> = \sin((90 -\alpha) - \beta) </math><math>= \sin (90- \alpha) \cos (\beta) - \sin ( \beta) \cos (90- \alpha) </math>
+Using the ratio definition of trigonometry, we apply [[Pythagorean Theorem]] on our triangle above to get 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 x + \cos^2 x =1 </math>. Using the unit circle definition of trigonometry, it's extremely easy to see that <math>\sin^2 x+ \cos^2 x =1 </math>.
-<math>=\cos \alpha \cos \beta - \sin \beta \sin \alpha </math>
+To derive the other two, divide by either <math>\sin^2 (x)</math> or <math>\cos^2 (x)</math> and substitute the respective trigonometry in place of the ratios.
-== Double Angle Identities ==
+== Angle Addition Identities ==
-Double angle identities are easily derived from the angle addition formulas by just letting <math> \alpha = \beta </math>.  Doing so yields:
+The trigonometric angle addition identities state the following identities:
+* <math>\sin(x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)</math>
+* <math>\cos(x + y) = \cos (x) \cos (y) - \sin (x) \sin (y) </math>
+* <math>\tan(x + y) = \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan (y)} </math>
+There are many proofs of these identities. For the sake of brevity, we list only one here.
-<cmath>\begin{eqnarray*}
+[[Euler's identity]] states that <math>e^{ix} = \cos (x) + i \sin(x)</math>. We have that
-\sin 2\alpha &=& 2\sin \alpha \cos \alpha\\
+<cmath>\begin{align*}
-\cos 2\alpha  &=& \cos^2 \alpha - \sin^2 \alpha\\
+\cos (x+y) + i \sin (x+y) &= e^{i(x+y)} \\
-&=& 2\cos^2 \alpha - 1\\
+&= e^{ix} \cdot e^{iy} \\
-&=& 1-2\sin^2 \alpha\\
+&= (\cos (x) + i \sin (x))(\cos (y) + i \sin (y)) \\
-\tan 2\alpha  &=& \frac{2\tan \alpha}{1-\tan^2\alpha} \end{eqnarray*}</cmath>
+&= (\cos (x) \cos (y) - \sin (x) \sin(y)) + i(\sin (x) \cos(y) + \cos(x) \sin(y))
+\end{align*}</cmath>
+By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas.
-=Further Conclusions=
+To derive the tangent addition formula, we reduce tangent to sine and cosine, divide both numerator and denominator by <math>\cos (x) \cos (y)</math>, and simplify.
+<cmath>\begin{align*}
+\tan (x+y) &= \frac{\sin (x+y)}{\cos (x+y)} \\
+&= \frac{\sin (x) \cos(y) + \cos(x) \sin(y)}{\cos (x) \cos (y) - \sin (x) \sin(y)} \\
+&= \frac{\frac{\sin(x)}{\cos(x)} + \frac{\sin(y)}{\cos(x)}}{1 - \frac{\sin (x) \sin(y)}{\cos (x) \cos(y)}} \\
+&= \frac{\tan (x) + \tan (y)}{1 - \tan (x) \tan(y)}
+\end{align*}</cmath>
+as desired.
-We can see from the above that
+== Double-Angle Identities ==
+The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting <math>x = y </math>.  Doing so yields:
+* <math>\sin (2x) = 2\sin (x) \cos (x)</math>
+* <math>\cos (2x) = \cos^2 (x) - \sin^2 (x)</math>
+* <math>\tan (2x) = \frac{2\tan (x)}{1-\tan^2 (x)}</math>
-*<math>\csc(2a) = \frac{\csc(a)\sec(a)}{2}</math>
+=== Cosine Double-Angle ===
-*<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>
+Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle identity given above.
-*<math>\cot(2a) = \frac{1 - \tan^2(a)}{2\tan(a)}</math>
+* <math>\cos (2x) = 1 - 2 \sin^2 (x)</math>
+* <math>\cos (2x) = 2 \cos^2 (x) - 1</math>
-== Half Angle Identities ==
+In addition, the following identities are useful in [[integration]] and in deriving the half-angle identities. They are a simple rearrangement of the two above.
-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.
+* <math>\sin^2 (x) = \frac{1 - \cos (2x)}{2}</math>
+* <math>\cos^2 (x) = \frac{1 + \cos (2x)}{2}</math>
-To summarize:
+== Half-Angle Identities ==
+The trigonometric half-angle identities state the following equalities:
+* <math>\sin (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{2}}</math>
+* <math>\cos (\frac{x}{2}) = \pm \sqrt{\frac{1 + \cos (x)}{2}}</math>
+* <math>\tan (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{1+\cos \theta}} = \frac{\sin x}{1 + \cos (x)} = \frac{1-\cos (x)}{\sin (X)}</math>
+The plus or minus is not saying that there are two answers, but that the sine of the angle depends on the quadrant that the angle is in.
-*<math> \sin \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}2} </math>
+Take the two expressions listed in the cosine double-angle section for <math>\sin^2 (x)</math> and <math>\cos^2 (x)</math>, and substitute <math>\frac{1}{2} x</math> instead of <math>x</math>. Taking the square root then yields the desired half-angle identities for sine and cosine.
-*<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}}=\frac{\sin \theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin \theta}  </math>
-== Prosthaphaeresis Identities ==
+== Sum-to-Product Identities ==
-(Otherwise known as sum-to-product identities)
 * <math>{\sin \theta + \sin \gamma = 2 \sin \frac{\theta + \gamma}2 \cos \frac{\theta - \gamma}2}</math>

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