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Circular Functions

A different name of an angle is circular functions. Communicate the direction of a triangle to the length of the surface of a triangle. Trigonometric functions are important of triangles and form episodic occurrence, along with many complementary applications. Trigonometric functions have a wide range of uses including calculating indefinite lengths along with angles in triangles.

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Six Circular Functions

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Trigonometric functions are normally specific as ratios of two sides of a right triangle including the angle, and able to equally specific as the lengths of different line segments from a unit circle.

Circular Function

More modern significance communicate them an infinite series or as solutions of specific different equations, allowing their extension to subjective positive also negative values and complex numbers. The sine with cosine functions are with usually used to model periodic function. Circular functions along angle $\theta$ are:

Sine Function:

sin $\theta$ = $\frac{\text{Opposite}}{\text{Hypotenuse}}$

Cosine Function:

cos $\theta$ = $\frac{\text{Adjacent}}{\text{Hypotenuse}}$

Tangent Function:

tan $\theta$ = $\frac{\text{Opposite}}{\text{Adjacent}}$

Cosecant Function:

csc $\theta$ = $\frac{\text{Hypotenuse}}{\text{Opposite}}$ = $\frac{1}{\sin \theta}$

Secant Function:

sec $\theta$ = $\frac{\text{Hypotenuse}}{\text{Adjacent}}$ = $\frac{1}{\cos \theta}$

Cotangent Function:

cot $\theta$ = $\frac{\text{Adjacent}}{\text{Opposite}}$ = $\frac{1}{\tan \theta}$

A function of a direction in a right-angled triangle to be specific the ratio of the length of the side opposed the angle to the length of the hypotenuse. A range of properties vary sinusoidal. They are able to be signifying diagrammatically by a sine wave.

Inverse Circular Functions

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If a function $f$ be a bijective (one-one and onto) from the set $A$ to $B$, then the inverse of $f$ is $f^{-1}: B \to A$, exist and is defined as the function which is associated each element $y$ belongs to $B$ to one and only one element $X$, belongs to $A$. Such that $y = f(x)$. Thus, if $y = f(x)$, then $x = f^{-1}(y)$. For the function $f$, $A$ is the domain and $B$ is the range, but for $f^{-1}$ the domain is $B$ and it's range is the set $A$. When this idea applied to circular functions gives inverse circular functions.

Inverse Circular Functions and Trigonometric Equations:


The inverse circular functions are defined as below:

1. sin-1 (-x) = - sin-1 x, -1 < x < 1 (Odd function)

2. cos-1 (-x) = $\pi$ - cos–1 x, -1 < x < 1 (Neither odd nor even)

3. tan-1(-x) = - tan-1 x, x ∈ R (Odd function)

4. cot-1 (-x) = $\pi$ - cot-1 x, x ∈ R (Neither odd nor even)

5. cosec-1 (-x) = -cosec-1 x, x < -1 or x > 1 (Odd function)

6. sec-1 (-x) = $\pi$ - sec-1 x, x < -1 or x > 1 (Neither odd nor even).

The inverse circular functions with their domain and range can be tabulated as:

Function
Domain
Range
Sin$^{-1}$x iff x = Sin $\theta$ [-1,1] [$\frac{-\pi}{2}$, $\frac{\pi}{2}$]
Cos$^{-1}$x iff x = Cos $\theta$ [-1,1] [0, $\pi$]
Tan$^{-1}$x iff x = Tan $\theta$ R ($\frac{-\pi}{2}$, $\frac{\pi}{2}$)
Cosec$^{-1}$x iff x = Cose $\theta$ R - (-1,1) [$\frac{-\pi}{2}$, $\frac{\pi}{2}$] - {0}
Sec$^{-1}$x iff x = Sec $\theta$ R - (-1,1) [0, $\pi$] - { $\frac{\pi}{2}$}
Cot$^{-1}$x iff x = Cot $\theta$ R (0, $\pi$)


Graphing Circular Functions

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Graphical representation of the circular functions are given below:

Graph for y = sin x



Graph for y = cos x



Graph for y = tan x


Circular Function Examples

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Given below are some of the problems based on trigonometric circular functions.

Solved Examples

Question 1:

The point $p(\frac{sqrt{3}}{2} , \frac{-1}{6})$ is on the unit circle. The length of the arc from the point A(1,0) to point p is q units. what are the values of the six circular functions of q.

Circular Function Examples


Solution:

From the above figure, sin and cosine functions are derived from the coordinate point p.

$\sin q$ = $\frac{-1}{6}$

$\cos q$ = $\frac{\sqrt{3}}{2}$

$\tan q$ = $\frac{\sin q}{\cos q}$ = $\frac{\frac{1}{6}}{\frac{\sqrt{3}}{2}}$

= $\frac{2}{6 \sqrt{3}}$

= $\frac{-1}{3 \sqrt{3}}$

$\cot q$ = $\frac{\cos q}{\sin q}$ = $\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{6}}$

= $\frac{-(6 \sqrt{3})}{2}$

= $3 \sqrt{3}$

$\sec q$ = $\frac{1}{\cos q}$ = $\frac{1}{\frac{\sqrt{3}}{2}}$ = $\frac{2}{\sqrt{3}}$

$\csc q$ = $\frac{1}{\sin q}$ = $-6$

Question 2: The point $p(\frac{1}{3} ,7)$ is on the unit circle the length of the arc from the point A(1,0) to point p is q units. what are the values of the six circular functions of q?
Solution:
Sin and cosine functions are derived from the coordinates point p.

$\sin q = 7$

$\cos q$ = $\frac{1}{3}$

$\tan q$ = $\frac{\sin q}{\cos q}$ = $\frac{7}{\frac{1}{3}}$ = $21$

$\cot q$ = $\frac{\cos q}{\sin q}$ = $\frac{\frac{1}{3}}{7}$ = $\frac{1}{21}$

$\sec q$ = $\frac{1}{\cos q}$ = $\frac{7}{\frac{1}{3}}$ = $21$

$\csc q$ = $\frac{1}{\sin q}$ = $\frac{1}{7}$

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