Solveeit Logo
Login

Question

Question: How do you solve \( \sin \theta = 0.8\, \) for \[90 < \theta < 180\]?...

How do you solve sin⁡θ=0.8 \sin \theta = 0.8\, for 90<θ<18090 < \theta < 180?

Explanation

Solution

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(0.8 ){\sin ^{ - 1}}\left( {0.8\,} \right) . Using the rule of allied angle the required solution in the second quadrant is θ=π−sin⁡−1(0.8 )\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⁡θ=0.8 \sin \theta = 0.8\, and we have to find its solution in the interval 90<θ<18090 < \theta < 180
sin⁡θ=0.8 \sin \theta = 0.8\,
Taking inverse of sine on both the sides of the equation, we get
sin⁡−1(sin⁡θ)=sin⁡−1(0.8 ){\sin ^{ - 1}}\left( {\sin \theta } \right) = {\sin ^{ - 1}}\left( {0.8\,} \right)
Since sin⁡−1(sin⁡x)=1{\sin ^{ - 1}}\left( {\sin x} \right) = 1 as they both are inverse of each other, we get
θ=sin⁡−1(0.8 )\theta = {\sin ^{ - 1}}\left( {0.8\,} \right)
The value of θ\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(0.8 ){\sin ^{ - 1}}\left( {0.8\,} \right) with the help of a calculator. We get sin⁡−1(0.8 )=53.13∘{\sin ^{ - 1}}\left( {0.8\,} \right) = {53.13^ \circ } .
But according to the question, θ\theta is given for 90<θ<18090 < \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⁡(π−x)\sin x = \sin \left( {\pi - x} \right) , we have
θ=π−sin⁡−1(0.8 ) θ=π−53.13∘   \theta = \pi - {\sin ^{ - 1}}\left( {0.8\,} \right) \\\ \theta = \pi - {53.13^ \circ } \;
The degree equivalent of π\pi radian is equal to 180∘{180^ \circ }
θ=180∘−53.13∘ θ=126.87∘   \theta = 18{0^ \circ } - {53.13^ \circ } \\\ \theta = 126.8{7^ \circ } \;
Therefore, the solution of the given trigonometric equation is θ=126.87∘\theta = 126.8{7^ \circ } , for 90<θ<18090 < \theta < 180.
So, the correct answer is “ θ=126.87∘\theta = 126.8{7^ \circ } , for 90<θ<18090 < \theta < 180.”.

Note : 1.Period of sine function is 2π2\pi
2. The domain of sine function is in the interval [−π2,π2]\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] and the range is in the interval [−1,1]\left[ { - 1,1} \right] .
3.Sine function is negative in 3rd and 4th quadrants.