Difference between revisions of "Trigonometric identities"

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== 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>

Revision as of 10:28, 4 August 2015

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:

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

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 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 $\tan A =\frac{\sin A}{\cos A}$.

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 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.

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

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 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:

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

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

  • $a^2 = b^2 + c^2 - 2bc\cos A.$

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

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

Other Identities

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

The two identities right above here were based on identites others posted on this site with a substitution.

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

See also