Question

How do you find the value of \[\sin 240{}^\circ \] ?

Answer 1

Hint: As we know that \[\sin \] function is a periodic function with its time period \[2\pi \] that means it repeats its output value after each interval of \[2\pi \]. Clearly \[240{}^\circ \] is in the third quadrant and also knows that the \[\sin \] function is negative in the third quadrant as it gives positive value only in the first and second quadrant.

Complete step-by-step answer:
Since \[\sin \] is a periodic function of time period \[2\pi \], also negative in third quadrant
\[\Rightarrow \sin (\pi +\theta )=-\sin \theta \] and \[\sin (\pi -\theta )=\sin \theta \] ,where \[\pi =180{}^\circ \]
Also, we can write \[\sin 240{}^\circ \text{ = sin(180}{}^\circ +30{}^\circ )\]
Using the property \[\sin (\pi +\theta )=-\sin \theta \]
Comparing the given question, here \[\theta =30{}^\circ \]
\[\Rightarrow \sin 240{}^\circ \]
\[\Rightarrow \sin (180{}^\circ +30{}^\circ )=-\sin 30{}^\circ \]
And we already know that \[\sin 30{}^\circ \text{ = }\dfrac{1}{2}\]
\[\Rightarrow \sin 240{}^\circ =-\dfrac{1}{2}\]

Note: When we have to calculate the value of trigonometric functions whose angles don’t lie in first quadrant then the first thing we have to remember is all functions are positive in first quadrant, \[\sin \] function positive in second quadrant while the \[\tan \] in third and \[\cos \] in fourth quadrant and \[\cos ec,\sec ,\cot \] are with their reciprocal functions. And when we modify the internal angle in terms of \[\pi \] and \[2\pi \] then the modulus value will be the same as those angle and the sign will be according the above rule and when modify in terms of \[\dfrac{\pi }{2}\] and \[\dfrac{3\pi }{2}\] then the \[\sin \] is replace by \[\cos \] , \[\tan \] is replaced by \[\cot \] and vice versa and that splitted angle will same.