
What are the 6 basic trigonometric functions?
Answer
514.8k+ views
Hint: The 6 basic trigonometric functions are derived from a right-angled triangle. Firstly we mark the Hypotenuse Base and perpendicular of the triangle then using them we derive the 3 basic trigonometric functions. Then by using these 3 trigonometric functions we derive the rest three of them.
Complete step by step answer:
The all 6 trigonometric functions are derived from a right-angled triangle.
Let the triangle be as below with each side labeled.
Where,
$a=$ Opposite side
$b=$ Adjacent side
$h=$ Hypotenuse
Now let us take $\angle C$ with respect to which we will find the value of all the basic trigonometric functions.
So starting with 3 Basic trigonometric function which one can find in the most calculators are below:
Sine:
$\sin C=$ Opposite/ Hypotenuse
$\sin C=\dfrac{a}{h}$…….$\left( 1 \right)$
Cosine:
$\cos C=$ Adjacent/Hypotenuse
$\cos C=\dfrac{b}{h}$…..$\left( 2 \right)$
Tangent:
$\tan C=$ Opposite/Adjacent
$\tan C=\dfrac{a}{b}$……$\left( 3 \right)$
Now we can find the rest 3 trigonometric function by the above 3 trigonometric function as:
Cosecant:
$\csc C=\dfrac{1}{\sin C}$
From equation (1) put the value in above equation,
$\Rightarrow \csc C=\dfrac{1}{\dfrac{a}{h}}$
$\therefore \csc C=\dfrac{h}{a}$…..$\left( 4 \right)$
Secant:
$\sec C=\dfrac{1}{\cos C}$
From equation (2) put the value in above equation,
$\Rightarrow \sec C=\dfrac{1}{\dfrac{b}{h}}$
$\therefore \sec C=\dfrac{h}{b}$…..$\left( 5 \right)$
Cotangent:
$\cot C=\dfrac{1}{\tan C}$
From equation (3) put the value in above equation,
$\Rightarrow \cot C=\dfrac{1}{\dfrac{a}{b}}$
$\therefore \cot C=\dfrac{b}{a}$…..$\left( 6 \right)$
So from equation (1)-(6) we get the basic trigonometric functions.
Hence 6 Basic trigonometric functions are sine, cosine, tangent, cosecant, secant and cotangent.
Note: Trigonometry is that branch of mathematics that studies the relation between the sides and the angle of triangles. Trigonometric function which is sine, cosine, tangent, cosecant, secant and cotangent helps in finding the missing side or missing angle of a triangle. There are various trigonometric identities used for calculation purposes.
Complete step by step answer:
The all 6 trigonometric functions are derived from a right-angled triangle.
Let the triangle be as below with each side labeled.
Where,
$a=$ Opposite side
$b=$ Adjacent side
$h=$ Hypotenuse
Now let us take $\angle C$ with respect to which we will find the value of all the basic trigonometric functions.
So starting with 3 Basic trigonometric function which one can find in the most calculators are below:
Sine:
$\sin C=$ Opposite/ Hypotenuse
$\sin C=\dfrac{a}{h}$…….$\left( 1 \right)$
Cosine:
$\cos C=$ Adjacent/Hypotenuse
$\cos C=\dfrac{b}{h}$…..$\left( 2 \right)$
Tangent:
$\tan C=$ Opposite/Adjacent
$\tan C=\dfrac{a}{b}$……$\left( 3 \right)$
Now we can find the rest 3 trigonometric function by the above 3 trigonometric function as:
Cosecant:
$\csc C=\dfrac{1}{\sin C}$
From equation (1) put the value in above equation,
$\Rightarrow \csc C=\dfrac{1}{\dfrac{a}{h}}$
$\therefore \csc C=\dfrac{h}{a}$…..$\left( 4 \right)$
Secant:
$\sec C=\dfrac{1}{\cos C}$
From equation (2) put the value in above equation,
$\Rightarrow \sec C=\dfrac{1}{\dfrac{b}{h}}$
$\therefore \sec C=\dfrac{h}{b}$…..$\left( 5 \right)$
Cotangent:
$\cot C=\dfrac{1}{\tan C}$
From equation (3) put the value in above equation,
$\Rightarrow \cot C=\dfrac{1}{\dfrac{a}{b}}$
$\therefore \cot C=\dfrac{b}{a}$…..$\left( 6 \right)$
So from equation (1)-(6) we get the basic trigonometric functions.
Hence 6 Basic trigonometric functions are sine, cosine, tangent, cosecant, secant and cotangent.
Note: Trigonometry is that branch of mathematics that studies the relation between the sides and the angle of triangles. Trigonometric function which is sine, cosine, tangent, cosecant, secant and cotangent helps in finding the missing side or missing angle of a triangle. There are various trigonometric identities used for calculation purposes.
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