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Question: What are the six trigonometric function values of 540?...

What are the six trigonometric function values of 540?

Explanation

Solution

Hint : To find the six trig function values of 540, we need to draw the unit circle first. The angle is given in degree so convert it into radian. Now, 540=3π540^\circ = 3\pi so sin3π\sin 3\pi and cos3π\cos 3\pi will lie on the X' axis. The values of sine and cosine on X’ axis are given by point (cosθ,sinθ)=(1,0)\left( {\cos \theta ,\sin \theta } \right) = \left( { - 1,0} \right) . Now, using these two values we can find the remaining 4 trig function values easily.

Complete step by step solution:
In this question, we have to find the six trigonometric function values of 540.
The six trigonometric functions are sin, cos, tan, cosec, sec and cot.
Now, the angle is given in degrees in our question. We have to convert it in radian first of all.
So, to convert degree into radian, we multiply degree with π180\dfrac{\pi }{{180}} .
Therefore, 540=540×π180=3π540^\circ = 540 \times \dfrac{\pi }{{180}} = 3\pi .Now, let us draw a unit circle to find these trig function values.

Now, we need to find sin3π,cos3π,tan3π,cosec3π,sec3π,cot3π\sin 3\pi ,\cos 3\pi ,\tan 3\pi ,\cos ec3\pi ,\sec 3\pi ,\cot 3\pi .

  1. sin3π\sin 3\pi
    All the points on the unit circle are represented by (cosθ,sinθ)\left( {\cos \theta ,\sin \theta } \right)
    3π3\pi lies on the X' axis and the coordinates at X’ axis are (1,0)\left( { - 1,0} \right) .
    Hence, the value of all sinθ\sin \theta on X’ axis will be equal to 0.
    Hence, sinπ=sin3π=sin5π=0\sin \pi = \sin 3\pi = \sin 5\pi = 0 .
    Therefore, sin3π=0\sin 3\pi = 0 .

  2. cos3π\cos 3\pi
    3π3\pi lies on the X' axis and the coordinates at X’ axis are (1,0)\left( { - 1,0} \right) .
    Hence, the value of all cosθ\cos \theta on X’ axis will be equal to 1- 1 .
    Hence, cosπ=cos3π=cos5π=1\cos \pi = \cos 3\pi = \cos 5\pi = - 1 .
    Therefore, cos3π=1\cos 3\pi = - 1 .

  3. tan3π\tan 3\pi
    We know that tanθ=sinθcosθ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} , and we have the values of sin3π\sin 3\pi and cos3π\cos 3\pi .
    Therefore, tan3π=sin3πcos3π\tan 3\pi = \dfrac{{\sin 3\pi }}{{\cos 3\pi }}
    tan3π=01 tan3π=0   \Rightarrow \tan 3\pi = \dfrac{0}{{ - 1}} \\\ \Rightarrow \tan 3\pi = 0 \;

  4. cosec3π\cos ec3\pi
    We know that cosec is the inverse of sine. So, we can write cosecθ\cos ec\theta as 1sinθ\dfrac{1}{{\sin \theta }} .
    cosec3π=1sin3π\Rightarrow \cos ec3\pi = \dfrac{1}{{\sin 3\pi }}
    And sin3π=0\sin 3\pi = 0
    cosec3π=10\Rightarrow \cos ec3\pi = \dfrac{1}{0}
    cosec3π=\Rightarrow \cos ec3\pi = not defined.

  5. sec3π\sec 3\pi
    We know that sec is the inverse of cos. So, we can write secθ\sec \theta as 1cosθ\dfrac{1}{{\cos \theta }} .
    sec3π=1cos3π\Rightarrow \sec 3\pi = \dfrac{1}{{\cos 3\pi }}
    And cos3π=1\cos 3\pi = - 1
    sec3π=11 sec3π=1   \Rightarrow \sec 3\pi = \dfrac{1}{{ - 1}} \\\ \Rightarrow \sec 3\pi = - 1 \;

  6. cot3π\cot 3\pi
    We know that cot is the inverse of tan. So, we can write cotθ\cot \theta as 1tanθ\dfrac{1}{{\tan \theta }} .
    cot3π1tan3π\Rightarrow \cot 3\pi \dfrac{1}{{\tan 3\pi }}
    And tan3π=0\tan 3\pi = 0 .
    cot3π=10\Rightarrow \cot 3\pi = \dfrac{1}{0}
    cot3π=\Rightarrow \cot 3\pi = not defined
    Hence, we have found all the six trig function values of 540.
    sin540=sin3π=0 cos540=cos3π=1 tan540=tan3π=0 cosec540=cosec3π=notdefined sec540=sec3π=1 cot540=cot3π=notdefined   \sin 540^\circ = \sin 3\pi = 0 \\\ \cos 540^\circ = \cos 3\pi = - 1 \\\ \tan 540^\circ = \tan 3\pi = 0 \\\ \cos ec540^\circ = \cos ec3\pi = notdefined \\\ \sec 540^\circ = \sec 3\pi = - 1 \\\ \cot 540^\circ = \cot 3\pi = notdefined \;

Note : In first quadrant (0θπ2)\left( {0 \leqslant \theta \leqslant \dfrac{\pi }{2}} \right) , all the values of trig functions are positive.
In second quadrant (π2θπ)\left( {\dfrac{\pi }{2} \leqslant \theta \leqslant \pi } \right) , the values of only sine and cosecant are positive
In third quadrant (πθ3π2)\left( {\pi \leqslant \theta \leqslant \dfrac{{3\pi }}{2}} \right) , the values of only tangent and cotangent are positive.
In fourth quadrant (3π2θ2π)\left( {\dfrac{{3\pi }}{2} \leqslant \theta \leqslant 2\pi } \right) , the values of only cosine and secant are positive.