Graphs of Trigonometric Functions
The relation (variation) between the angles and the values of the trigonometric ratios at them are plotted by graphs .
Pro 1: Graph of y = sinx
x -`pi` `-pi/2` 0 `pi/2` `pi` `(3pi)/2` `2pi` `(5pi)/2` `3pi`
y 0 -1 0 1 0 -1 0 1 0
Plot the graph in the coordinate plane by taking angles x in radian measure on X-axis and the values of sinx = y on Y-axis .
Sol :
By choosing a suitable scale , plot and join the points of y = sin x with a smooth curve to get the graph .
This curve passes through the origin . The values of sin x vary between -1 and +1 which are respectively the minimum and the maximum . It is in the shape of a wave whose wave length is 2`pi` . This wavelength is nothing but the period .
SInce `-1<=sinx<=1` `AA` x `in` R , the sine function is bounded . It can be proved that it is a continuous function on R .
Pro 2: Graph of y = cosx
x `-pi` -`pi/2` 0 `pi/2` `pi` `(3pi)/2` `2pi` `(5pi)/2`
y -1 0 1 0 -1 0 1 0
Sol :
By choosing a suitable scale , plot and join the points of y = cos x with a smooth curve to get the graph .
This curve does not pass through the origin . It is evident that the maximum and the minimum values are +1 and -1 respectively .
Since `-1<=cosx<=1` `AA` x `in` R , the cosine function is bounded . It can be shown that is a contnous function and periodic with `2pi` as the period .
I like to share this Solving Trigonometric Equations with you all through my article.
Pro 3: Graph of y = tanx
x `-pi/2` 0 `pi/2` `pi` `(3pi)/2`
y not defined 0 not defined 0 not defined
Sol :
The curve nearly touches the vertical lines at x = ...........`-pi/2` , `pi/2` , `(3pi)/2` , . .. . . . .
The curve has nreakes at x = (2n + 1)`pi/2` , n `in` Z and passes through the origin . it is not bounded .
The tan function is periodic and `pi` is the period of it .
Pro 4: Graph of y = cotx
The curve nearly touches the vertical lines at x = . . . . . . . `-pi` , 0 , `pi` . . . . . . . .
The curve has breakes at x = n`pi` , n `in` Z and does not pass through the origin . It is not bounded .
The cot function is periodic and `pi` is the period of it .
Having problem with Trig Identities Solver Read my upcoming post, i will try to help you.
Solve Graphing Trigonometric Functions : Secx and Cosecx
Pro 5: Graph of y = secx
The curve nearly touches the vertical lines at x = . . . . . . . . `-pi/2` , 0 , `pi/2` , `pi` , `(3pi)/2` . . . . . .
The curve has breakes at x `in` (2n+1) `pi/2` , n `in` Z . It is not bounded . The values of secx lie in `(oo,-1]uu[1,oo)`
The secant function is periodic and 2`pi` is the period of it .
Understanding find answers to math problems is always challenging for me but thanks to all math help websites to help me out.
Pro 6: Graph of y = cosecx
The curve nearly touches the vertical lines at x = . . . . . . 0 , `pi/2` , `pi` , `(3pi)/2` , 2`pi` , . . . . . .
The curve has taken
The relation (variation) between the angles and the values of the trigonometric ratios at them are plotted by graphs .
Pro 1: Graph of y = sinx
x -`pi` `-pi/2` 0 `pi/2` `pi` `(3pi)/2` `2pi` `(5pi)/2` `3pi`
y 0 -1 0 1 0 -1 0 1 0
Plot the graph in the coordinate plane by taking angles x in radian measure on X-axis and the values of sinx = y on Y-axis .
Sol :
By choosing a suitable scale , plot and join the points of y = sin x with a smooth curve to get the graph .
This curve passes through the origin . The values of sin x vary between -1 and +1 which are respectively the minimum and the maximum . It is in the shape of a wave whose wave length is 2`pi` . This wavelength is nothing but the period .
SInce `-1<=sinx<=1` `AA` x `in` R , the sine function is bounded . It can be proved that it is a continuous function on R .
Pro 2: Graph of y = cosx
x `-pi` -`pi/2` 0 `pi/2` `pi` `(3pi)/2` `2pi` `(5pi)/2`
y -1 0 1 0 -1 0 1 0
Sol :
By choosing a suitable scale , plot and join the points of y = cos x with a smooth curve to get the graph .
This curve does not pass through the origin . It is evident that the maximum and the minimum values are +1 and -1 respectively .
Since `-1<=cosx<=1` `AA` x `in` R , the cosine function is bounded . It can be shown that is a contnous function and periodic with `2pi` as the period .
I like to share this Solving Trigonometric Equations with you all through my article.
Solve Graphing Trigonometric Functions : Tanx and Cotx
Pro 3: Graph of y = tanx
x `-pi/2` 0 `pi/2` `pi` `(3pi)/2`
y not defined 0 not defined 0 not defined
Sol :
The curve nearly touches the vertical lines at x = ...........`-pi/2` , `pi/2` , `(3pi)/2` , . .. . . . .
The curve has nreakes at x = (2n + 1)`pi/2` , n `in` Z and passes through the origin . it is not bounded .
The tan function is periodic and `pi` is the period of it .
Pro 4: Graph of y = cotx
The curve nearly touches the vertical lines at x = . . . . . . . `-pi` , 0 , `pi` . . . . . . . .
The curve has breakes at x = n`pi` , n `in` Z and does not pass through the origin . It is not bounded .
The cot function is periodic and `pi` is the period of it .
Having problem with Trig Identities Solver Read my upcoming post, i will try to help you.
Solve Graphing Trigonometric Functions : Secx and Cosecx
Pro 5: Graph of y = secx
The curve nearly touches the vertical lines at x = . . . . . . . . `-pi/2` , 0 , `pi/2` , `pi` , `(3pi)/2` . . . . . .
The curve has breakes at x `in` (2n+1) `pi/2` , n `in` Z . It is not bounded . The values of secx lie in `(oo,-1]uu[1,oo)`
The secant function is periodic and 2`pi` is the period of it .
Understanding find answers to math problems is always challenging for me but thanks to all math help websites to help me out.
Pro 6: Graph of y = cosecx
The curve nearly touches the vertical lines at x = . . . . . . 0 , `pi/2` , `pi` , `(3pi)/2` , 2`pi` , . . . . . .
The curve has taken
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