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Question

Question: How do you graph y = tan3x?...

How do you graph y = tan3x?

Explanation

Solution

If we need to find the values at different angles, we apply trigonometric ratios. Tangent function is formed by dividing sine and cosine function. Also, we should know about the even and odd functions. In this question, we should know about basic function tanx and we have drawn a graph of tan3x.

Complete step by step answer:
Some basic trigonometric functions are:
\Rightarrow Sine (sin)
\Rightarrow Cosine (cos)
\Rightarrow Tangent (tan)
So, when we say tanθ\tan \theta , here θ\theta means angle in degrees.
Derived functions are derived from the basic trigonometric functions. They are as follows:
\Rightarrow cosecθ\theta = 1sinθ\dfrac{1}{\sin \theta }
\Rightarrow secθ\theta = 1cosθ\dfrac{1}{\cos \theta }
\Rightarrow tanθ\theta = sinθcosθ\dfrac{\sin \theta }{\cos \theta } = 1cotθ\dfrac{1}{\cot \theta }
\Rightarrow cotθ\theta = 1tanθ\dfrac{1}{\tan \theta } = cosθsinθ\dfrac{\cos \theta }{\sin \theta }
You know what exactly tanθ\theta is? Let’s find it out.

So, from above figure,
\Rightarrow sinθ\theta = perpendicular(P)hypotenuse(H)\dfrac{perpendicular(P)}{hypotenuse(H)}
\Rightarrow cosθ\theta = base(B)hypotenuse(H)\dfrac{base(B)}{hypotenuse(H)}
tanθ=sinθcosθ tanθ=perpendicular(P)base(B) \begin{aligned} & \Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta } \\\ & \Rightarrow \tan \theta =\dfrac{perpendicular(P)}{base(B)} \\\ \end{aligned}
Now, let’s see some even and odd functions.
\Rightarrow sin(-x) = -sinx
\Rightarrow cos(-x) = cosx
\Rightarrow tan(-x) = -tanx
\Rightarrow cot(-x) = -cotx
\Rightarrow cosec(-x) = -cosecx
\Rightarrow sec(-x) = secx
Now, let’s make a table of trigonometric ratios for basic trigonometric functions i.e. sin, cos, tan, cot, sec and cosec.

Trigonometric ratios(angle θ\theta in degrees)0{{0}^{\circ }}30{{30}^{\circ }}45{{45}^{\circ }}60{{60}^{\circ }}90{{90}^{\circ }}
sinθ\theta 012\dfrac{1}{2}12\dfrac{1}{\sqrt{2}}32\dfrac{\sqrt{3}}{2}1
cosθ\theta 132\dfrac{\sqrt{3}}{2}12\dfrac{1}{\sqrt{2}}12\dfrac{1}{2}0
tanθ\theta 013\dfrac{1}{\sqrt{3}}13\sqrt{3}\infty
cosecθ\theta \infty 22\sqrt{2}23\dfrac{2}{\sqrt{3}}1
secθ\theta 123\dfrac{2}{\sqrt{3}}2\sqrt{2}2\infty
cotθ\theta \infty 3\sqrt{3}113\dfrac{1}{\sqrt{3}}0

Graph of tan3x is quite similar to tanx.
As period of y = tan3x so, tan3x takes all the values with interval length of π3\dfrac{\pi }{3}.
First, let’s see the graph of tanx.

Now, let’s see the graph of tan3x.

Note: Students should remember all the functions and trigonometric ratios before solving any question related to trigonometry. In this question, the main purpose of the question is to sketch the graph of tan3x which should be done neatly.