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$$\text{Range}=[-\mathrm{A,A}]$$
We now consider yet another form of sine function which is given as :
$$f\left(x\right)=A\mathrm{sin}\left(kx\right)$$
Multiplying argument x of sine function by a constant k does not change the nature of plot. However, it changes the periodicity of the function. Recall that if T is the period of function f(x), then period of function $af(kx\pm b)$ is $\frac{T}{\left|k\right|}$ Clearly, the period of sin(kx) is $\frac{T}{\left|k\right|}$ . If |k| is less than 1, then period is more than 2π and if |k| is greater than 1, then period is less than 2π.
Problem : Find domain and range of function :
$$f\left(x\right)=\mathrm{sin}x+2$$
Solution : We know that domain of sinx is real number set R and range is [-1,1]. The given function is real for all real values of x. Hence, its domain remains R. On the other hand, minimum and maximum values of function changes from that corresponding to sinx function :
$${y}_{\text{min}}=-1+2=1$$ $${Y}_{\mathrm{max}}=1+2=3$$
Hence, range of given function is [1,3]. It is evident that graph of function is that of graph of sinx shifted up by 2 units.
For each real number “x”, there is a cosine function defined as :
$$f\left(x\right)=\mathrm{cos}\left(x\right)$$
The plot of cos(x) .vs. x is shown here.
The plot, here, is continuous and period is "2π". Think period of the function in term of minimum segment which can be used to extend the plot on either side. Further as cos(-x) = cosx, cosine function is an even function. This fact is also substantiated by the fact that plot is symmetric about y-axis.
Since function holds for all values of “x”, its domain is “R”. On the other hand, the values of cosine function is bounded between “-1” and “1”, inclusive of end points. Hence, domain and range of sine function are :
$$\text{Domain}=R$$
$$\text{Range}=[-\mathrm{1,1}]$$
When cosine function is given as f(x) = Acosx, maximum and minimum values of function becomes -A and A. The range is modified as :
$$\text{Range}=[-\mathrm{A,A}]$$
When cosine function is given as f(x) = Acos(kx), the period of cosine function is given by $\frac{T}{\left|k\right|}$ .
Problem : Find domain range of the function :
$$f\left(x\right)=12\mathrm{sin}x+5\mathrm{cos}x$$
Solution : The given function comprises of sine and cosine functions. Here, we reduce given function in terms of one trigonometric function and then find range of the function. This reduction is required as otherwise it would be difficult to estimate when two trigonometric functions together evaluates to minimum and maximum values. Let us put,
$$a\mathrm{cos}\alpha =12$$ $$a\mathrm{sin}\alpha =5$$
Clearly, $a=\sqrt{{12}^{2}+{5}^{2}}=13$ . Putting these values/ expression in function,
$$f\left(x\right)=13\left(\mathrm{cos}\alpha \mathrm{sin}x+\mathrm{sin}\alpha \mathrm{cos}x\right)=13\mathrm{sin}\left(x+\alpha \right)$$
We know that range of sine function is [-1,1]. Hence, range of given function is :
$$\text{Range}[-13,13]$$
For a real number “x”, there is a tangent function defined as :
$$f\left(x\right)=\mathrm{tan}\left(x\right)$$
Note that defining statement defines the function for a real number “x” – not for "each" real “x” as in the case of sine and cosine functions. It is so because, tangent function is not defined for all real values of “x”. Let us recall that :
$$\Rightarrow \mathrm{tan}x=\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}$$
This is a rational polynomial form, which is defined for $\mathrm{cos}\left(x\right)\ne 0$ . Now, cos(x) evaluates to zero for certain values of “x”, which appears at a certain interval given by the condition,
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