Trigonometric identities

Revision as of 18:16, 25 October 2007 by Temperal (talk | contribs) (lists not tables 3)

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:

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:

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

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)$$= \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:

$\sin 2\alpha =$2\sin \alpha \cos \alpha$$ (Error compiling LaTeX. Unknown error_msg)\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$$ (Error compiling LaTeX. Unknown error_msg)=2\cos^2 \alpha - 1$$ (Error compiling LaTeX. Unknown error_msg)=1-2\sin^2 \alpha$$ (Error compiling LaTeX. Unknown error_msg)\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:$ (Error compiling LaTeX. Unknown error_msg) \sin \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}2} $$ (Error compiling LaTeX. Unknown error_msg) \cos \frac{\theta}2 = \pm \sqrt{\frac{1+\cos \theta}2} $$ (Error compiling LaTeX. Unknown error_msg) \tan \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} $== Even-Odd Identities ==$\sin (-\theta) = -\sin (\theta) $$ (Error compiling LaTeX. Unknown error_msg)\cos (-\theta) = \cos (\theta) $$ (Error compiling LaTeX. Unknown error_msg)\tan (-\theta) = -\tan (\theta) $$ (Error compiling LaTeX. Unknown error_msg)\csc (-\theta) = -\csc (\theta) $$ (Error compiling LaTeX. Unknown error_msg)\sec (-\theta) = \sec (\theta) $$ (Error compiling LaTeX. Unknown error_msg)\cot (-\theta) = -\cot (\theta) $==Prosthaphaeresis Identities== (Otherwise known as sum-to-product identities)

  • $ (Error compiling LaTeX. Unknown error_msg)\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|Law of Sines}} The extended [[Law of Sines]] states

  • $ (Error compiling LaTeX. Unknown error_msg)\frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C} = 2R.$== Law of Cosines ==

{{main|Law of Cosines}} The [[Law of Cosines]] states

  • $ (Error compiling LaTeX. Unknown error_msg)a^2 = b^2 + c^2 - 2bc\cos A. $== Law of Tangents ==

{{main|Law of Tangents}} The [[Law of Tangents]] states

  • $ (Error compiling LaTeX. Unknown error_msg)\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}$

See also