How do you solve $ \sin (x) = 0.5 $ ?
Answer
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Hint: In order to determine the value of the above question, convert the decimal into fraction and use the trigonometric table to find the angle which is in the interval $ \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] $ whose sine is\[\dfrac{1}{2}\] to get the required result.
Complete step-by-step answer:
Given,
$ \sin (x) = 0.5 $
Write decimal value into the fraction
$
\sin (x) = 0.5 \\
\sin (x) = \dfrac{5}{{10}} \\
\sin (x) = \dfrac{1}{2} \;
$
Transposing sine from LHS to RHS
$ x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) $
We know that $ {\sin ^{ - 1}}\theta $ denotes an angle in the interval $ \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] $ whose sine is $ x $ for $ x \in \left[ { - 1,1} \right]. $
Therefore,
$ x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) $ = An angle in $ \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] $ , whose sine is $ \dfrac{1}{2} $ .
From the trigonometric table we have,
$ \sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2} $
Transposing sin from left-hand side to right-hand side
$ x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6} $
Therefore, the value of x is equal to $ \dfrac{\pi }{6} $ .
So, the correct answer is “$ \dfrac{\pi }{6} $”.
Note: In inverse trigonometric function, the domain are the ranges of corresponding trigonometric functions and the range are the domain of the corresponding trigonometric function.
3. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
4. Periodic Function= A function $ f(x) $ is said to be a periodic function if there exists a real number T > 0 such that $ f(x + T) = f(x) $ for all x.
Complete step-by-step answer:
Given,
$ \sin (x) = 0.5 $
Write decimal value into the fraction
$
\sin (x) = 0.5 \\
\sin (x) = \dfrac{5}{{10}} \\
\sin (x) = \dfrac{1}{2} \;
$
Transposing sine from LHS to RHS
$ x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) $
We know that $ {\sin ^{ - 1}}\theta $ denotes an angle in the interval $ \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] $ whose sine is $ x $ for $ x \in \left[ { - 1,1} \right]. $
Therefore,
$ x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) $ = An angle in $ \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] $ , whose sine is $ \dfrac{1}{2} $ .
From the trigonometric table we have,
$ \sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2} $
Transposing sin from left-hand side to right-hand side
$ x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6} $
Therefore, the value of x is equal to $ \dfrac{\pi }{6} $ .
So, the correct answer is “$ \dfrac{\pi }{6} $”.
Note: In inverse trigonometric function, the domain are the ranges of corresponding trigonometric functions and the range are the domain of the corresponding trigonometric function.
3. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
4. Periodic Function= A function $ f(x) $ is said to be a periodic function if there exists a real number T > 0 such that $ f(x + T) = f(x) $ for all x.
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