Question

How do you solve $ \sin \theta = 0.8\, $ for \[90 < \theta < 180\]?

Answer 1

Hint: In order to determine the solution of the above trigonometric equation, take inverse of sine on both sides. Since the values 0.8 is not a remarkable value for the sine function. Take use of the calculator to find the value of $ {\sin ^{ - 1}}\left( {0.8\,} \right) $ . Using the rule of allied angle the required solution in the second quadrant is $ \theta = \pi - {\sin ^{ - 1}}\left( {0.8\,} \right) $ . Simplify to obtain the required solution.

Complete step by step solution:
We are given a trigonometric equation $ \sin \theta = 0.8\, $ and we have to find its solution in the interval \[90 < \theta < 180\]
 $ \sin \theta = 0.8\, $
Taking inverse of sine on both the sides of the equation, we get
 $ {\sin ^{ - 1}}\left( {\sin \theta } \right) = {\sin ^{ - 1}}\left( {0.8\,} \right) $
Since $ {\sin ^{ - 1}}\left( {\sin x} \right) = 1 $ as they both are inverse of each other, we get
 $ \theta = {\sin ^{ - 1}}\left( {0.8\,} \right) $
The value of $ \theta $ is an angle whose sine value is equal to 0.8. As we can clearly see 0.8 is not a remarkable value for the sine function. So calculate the value of $ {\sin ^{ - 1}}\left( {0.8\,} \right) $ with the help of a calculator. We get $ {\sin ^{ - 1}}\left( {0.8\,} \right) = {53.13^ \circ } $ .
But according to the question, $ \theta $ is given for \[90 < \theta < 180\]. Sine function is positive in the 1st and 2nd quadrant. Using allied angle property for the angle in the second quadrant $ \sin x = \sin \left( {\pi - x} \right) $ , we have
 $
  \theta = \pi - {\sin ^{ - 1}}\left( {0.8\,} \right) \\
  \theta = \pi - {53.13^ \circ } \;
  $
The degree equivalent of $ \pi $ radian is equal to $ {180^ \circ } $
 $
  \theta = 18{0^ \circ } - {53.13^ \circ } \\
  \theta = 126.8{7^ \circ } \;
  $
Therefore, the solution of the given trigonometric equation is $ \theta = 126.8{7^ \circ } $ , for \[90 < \theta < 180\].
So, the correct answer is “ $ \theta = 126.8{7^ \circ } $ , for \[90 < \theta < 180\].”.

Note: 1.Period of sine function is $ 2\pi $
2. The domain of sine function is in the interval $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ and the range is in the interval $ \left[ { - 1,1} \right] $ .
3.Sine function is negative in 3rd and 4th quadrants.