
How do you evaluate $sin120^ \circ$
Answer
559.8k+ views
Hint: Hint:
The above problem is based on the calculation of the value of sinusoidal function when angle is given to us.
$sin120^ \circ$ is equal to $\dfrac{{\sqrt 3 }}{2}$ the value is positive because it lies in the second quadrant.
By using the above value we will further elaborate the problem.
Complete step-by-step answer:
Sign of the sinusoidal function will depend on the quadrant in which they exist. We have four quadrants in the x – y plane. Upper First quadrant from the right hand side is the quadrant in which all the values be it cosine, tangent, sine are positive and contains values from 0 to 90 degree. Then, comes the second quadrant, second quadrant in the upper half of the plane has only sine –cosec values positive and the angles are from 90 to 180 degree. The third quadrant lower quadrant of the LHS in the plane angles is from 180 to 270 degree and only tangent – cot values are positive. Fourth quadrant has cosine-secant values that are positive which contain angles from 270 to 360 degrees.
There, 120 degree lies in the second quadrant with $sin120^ \circ$ which gives us positive value and the magnitude is;
$ \Rightarrow \sin {120^\circ} = \dfrac{{\sqrt 3 }}{2}$
Note:
Like the $sin120^ \circ$ is equal to $\dfrac{{\sqrt 3 }}{2}$ , in the similar manner sin300 is equal to 1/2, $sin60^ \circ$ is equal to $\dfrac{{\sqrt 3 }}{2}$, $sin90^ \circ$ is equal to 1. Cosine too have different values the way sin has, $cos30^ \circ$ is equal to $\dfrac{{\sqrt 3 }}{2}$, $cos60^ \circ$ is equal to 1/2, $cos45^ \circ$ is equal to $\dfrac{1}{{\sqrt 2 }}$, $cos90^ \circ$ is equal to zero.
The above problem is based on the calculation of the value of sinusoidal function when angle is given to us.
$sin120^ \circ$ is equal to $\dfrac{{\sqrt 3 }}{2}$ the value is positive because it lies in the second quadrant.
By using the above value we will further elaborate the problem.
Complete step-by-step answer:
Sign of the sinusoidal function will depend on the quadrant in which they exist. We have four quadrants in the x – y plane. Upper First quadrant from the right hand side is the quadrant in which all the values be it cosine, tangent, sine are positive and contains values from 0 to 90 degree. Then, comes the second quadrant, second quadrant in the upper half of the plane has only sine –cosec values positive and the angles are from 90 to 180 degree. The third quadrant lower quadrant of the LHS in the plane angles is from 180 to 270 degree and only tangent – cot values are positive. Fourth quadrant has cosine-secant values that are positive which contain angles from 270 to 360 degrees.
There, 120 degree lies in the second quadrant with $sin120^ \circ$ which gives us positive value and the magnitude is;
$ \Rightarrow \sin {120^\circ} = \dfrac{{\sqrt 3 }}{2}$
Note:
Like the $sin120^ \circ$ is equal to $\dfrac{{\sqrt 3 }}{2}$ , in the similar manner sin300 is equal to 1/2, $sin60^ \circ$ is equal to $\dfrac{{\sqrt 3 }}{2}$, $sin90^ \circ$ is equal to 1. Cosine too have different values the way sin has, $cos30^ \circ$ is equal to $\dfrac{{\sqrt 3 }}{2}$, $cos60^ \circ$ is equal to 1/2, $cos45^ \circ$ is equal to $\dfrac{1}{{\sqrt 2 }}$, $cos90^ \circ$ is equal to zero.
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