Trigonometric Even-odd Functions

Even and odd functionsare functions satisfying certainsymmetries: even functions satisfy $f(x)=f(-x)$for all $x$, while odd functions satisfy $f(x)=-f(-x)$. Trigonometric functions are examples of non-polynomialeven (in the case of cosine) and odd (in the case of sine and tangent) functions.

The properties of even and odd functions are useful in analyzing trigonometric functions, particularly in thesum and difference formulas.

The cosine and sine functions satisfy the following properties:

$\begin{aligned} \cos(-\theta) &= \cos \theta \\ \sin(-\theta) &= -\sin \theta. \end{aligned}$

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Proof:
From the definition of cosine and sine in the unit circle,

$x= \cos \theta \quad \text{ and } \quad y= \sin \theta.$

We can see that for both $\theta$and $-\theta$, $x$remains the same. Thus, $\cos \theta=\cos (-\theta)$.

Similarly, we can see that the $y$in two cases are additive inverse of each other. Thus, $\sin (-\theta)=-\sin\theta$. $_\square$

Now that we have the above identities, we can prove several other identities, as shown in the following example:

Prove the identities

$\begin{aligned} \tan(-\theta) &= -\tan \theta\\ \cot(-\theta) &= -\cot \theta\\ \csc(-\theta) &= -\csc \theta\\ \sec(-\theta) &= \sec \theta. \end{aligned}$

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We have

$\begin{aligned} \tan(-\theta) &=\frac{\sin(-\theta)}{\cos(-\theta)}=\frac{-\sin \theta}{\cos \theta}=-\tan \theta\\ \cot(-\theta) &=\frac{1}{\tan(-\theta)}=\frac{1}{-\tan \theta}=-\cot \theta\\ \csc(-\theta) &=\frac{1}{\sin (-\theta)}=\frac{1}{-\sin \theta}=-\csc \theta\\ \sec(-\theta) &=\frac{1}{\cos(-\theta)}=\frac{1}{\cos \theta}=\sec \theta. \ _\square \end{aligned}$

Evaluate $\sin(-60^\circ) .$

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From trigonometric even-odd functions, we have

$\begin{aligned} \sin (-60^\circ) &= -\sin 60^\circ \\ &= -\frac{\sqrt{3}}{2}. \ _\square \end{aligned}$

Evaluate $\tan(-45^\circ).$

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From trigonometric even-odd functions, we have

$\begin{aligned} \tan (-45^\circ) &= -\tan 45^\circ \\ &= -1. \ _\square \end{aligned}$

Evaluate $\cos(-30^\circ)+\sec(-60^\circ).$

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From trigonometric even-odd functions, we have

$\begin{aligned} \cos(-30^\circ)+\sec(-60^\circ) &= \cos 30^\circ + \sec 60^\circ \\ &= \frac{\sqrt{3}}{2} + \frac{1}{\cos 60^\circ} \\ &= \frac{\sqrt{3}}{2} + 2 \\ &= \frac{\sqrt{3} + 4}{2}. \ _\square \end{aligned}$

Simplify $\tan x \times \cot(-x).$

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From trigonometric even-odd functions, we have

$\begin{aligned} \tan x \times \cot(-x) &= \tan x \times (-\cot x ) \\ &= \tan x \times \left(-\frac{1}{\tan x}\right) \\ &= -1. \ _\square \end{aligned}$

ParallelAngleBisector

Rectangle $ABCD$is inscribed in circle $O$with radius 1 in the above diagram. Let $\theta$be the angle formed by $\overline {AO}$and the $x$-axis. Then which vertex has

$2\sin\left(-\frac{\theta}{2}\right)\cos\left(-\frac{\theta}{2}\right)$

as its $y$-coordinate?

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Observe that the expression can be rewritten as follows:

$\begin{aligned} 2\sin\left(-\frac{\theta}{2}\right)\cos\left(-\frac{\theta}{2}\right) &= -2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} &&\quad \big(\text{since } \sin (-x) = -\sin x, \cos(-x) = \cos x\big) \\ &= -\sin \theta. &&\quad (\text{since } \sin 2\theta = 2\sin x \cos x ) \end{aligned}$

Also observe that the $y$-coordinate of point $A$is $\sin \theta$and point $A$and $C$are symmetric with respect to the origin. Then the $y$-coordinate of point $C$must be $-\sin \theta.$

Therefore, our answer is vertex $C.$ $_\square$

• Trigonometry
• 偶数和奇数的功能