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Question: How do you solve \(\sin \theta = - 0.5\)...

How do you solve sin⁡θ=−0.5\sin \theta = - 0.5

Explanation

Solution

We will simplify the given equation and then by using the trigonometric table, we will find the value of xx. Finally we get the required answer.

Complete step-by-step solution:
The given term is: sin⁡θ=−0.5\sin \theta = - 0.5
Now we know that the number −0.5 - 0.5 can be written in the form of a fraction as −12 - \dfrac{1}{2} therefore, on substituting it in the right-hand side of the expression, we get:
⇒sin⁡θ=−12\Rightarrow \sin \theta = - \dfrac{1}{2}
Now from the trigonometric table, we know that:
⇒sin⁡(−π6)=(sin⁡(π−(−π6)))\Rightarrow \sin \left( { - \dfrac{\pi }{6}} \right) = \left( {\sin \left( {\pi - \left( { - \dfrac{\pi }{6}} \right)} \right)} \right)
This can be written as:
⇒sin⁡(7π6)\Rightarrow \sin \left( {\dfrac{{7\pi }}{6}} \right)
This has the value: 12\dfrac{1}{2}
Therefore,
⇒x=(π6)c\Rightarrow x = {\left( {\dfrac{\pi }{6}} \right)^c}which is 30∘{30^\circ } , x=(7π6)cx = {\left( {\dfrac{{7\pi }}{6}} \right)^c}which is 210∘{210^\circ }
Now we know that sin⁡(−θ)=−sin⁡θ\sin ( - \theta ) = - \sin \theta

Therefore, on generalizing the answer, we get:
⇒x=(2nπ−π6)or(2nπ+7π6),n→ε→Z\Rightarrow x = \left( {2n\pi - \dfrac{\pi }{6}} \right)or\left( {2n\pi + \dfrac{{7\pi }}{6}} \right),n \to \varepsilon \to \mathbb{Z} , which is the required answer.

Note: This question can also be done by using the inverse trigonometric function as:
We have the given equation after simplification as: sin⁡θ=−12\sin \theta = - \dfrac{1}{2}
Now using the inverse trigonometric function, we get:
⇒θ=sin⁡−1(−12)\Rightarrow \theta = {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)
Therefore, the principal value of sin⁡x\sin x is π6\dfrac{\pi }{6}
Now we know that sin⁡(−θ)=−sin⁡θ\sin ( - \theta ) = - \sin \theta
Therefore, the principal value becomes −π6 - \dfrac{\pi }{6}
Now since sine is positive in the first and second quadrant, we will subtract the principal value from π\pi to get the solution in the second quadrant.
Therefore,
⇒π−−(π6)\Rightarrow \pi - - \left( {\dfrac{\pi }{6}} \right)
On simplifying we get:
⇒7π6\Rightarrow \dfrac{{7\pi }}{6}, which is the solution in the second quadrant.
Therefore, on generalizing the answer we get:
⇒x=(2nπ−π6)or(2nπ+7π6),n→ε→Z\Rightarrow x = \left( {2n\pi - \dfrac{\pi }{6}} \right)or\left( {2n\pi + \dfrac{{7\pi }}{6}} \right),n \to \varepsilon \to \mathbb{Z}, which is the required answer.
It is to be remembered which trigonometric functions are positive and negative in what quadrants.
The formula used over here is for sin⁡(nπ+x)\sin (n\pi + x) ,
It is to be remembered that sin⁡(nπ+x)=(−1)nsin⁡x\sin (n\pi + x) = {( - 1)^n}\sin x
Basic trigonometric formulas should be remembered to solve these types of sums.
The inverse trigonometric function of sin⁡x\sin x which is sin⁡−1x{\sin ^{ - 1}}x used in this sum
For example, if sin⁡x=a\sin x = a then x=sin⁡−1ax = {\sin ^{ - 1}}a .
And sin⁡−1(sin⁡x)=x{\sin ^{ - 1}}(\sin x) = x is a property of the inverse function.
There also exists inverse function for the other trigonometric relations such as tan and cos.
The inverse function is used to find the angle xx from the value of the trigonometric relation.