Question

How do you find the value of $\sin \left( {112\dfrac{1}{2}} \right)$ using the double or half angle formula.

Answer 1

Hint : Here we have to find the value of the sine trigonometry ratio. In the question it’s already mentioned that we have to solve the above function by using the double angle or half angle formula. By using the table of trigonometry ratios we can determine the value for the given question.

Complete step by step solution:
The concept known as a double angle is associated with the three common trigonometric ratios: sine (sin), cosine (cos), and tangent (tan). Double, as the word implies, means to increase the size of the angle to twice its size.
The double angle and the half angle formula is defined as $\cos 2\theta = 1 - 2{\sin ^2}\theta$ and ${\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$
Here the value of theta $\theta = 112\dfrac{1}{2}$ is in the degree.
Now consider the formula $\cos 2\theta = 1 - 2{\sin ^2}\theta$
Substituting the value of $\theta$ then we get
$\Rightarrow \cos 2\left( {112\dfrac{1}{2}} \right) = 1 - 2{\sin ^2}\left( {112\dfrac{1}{2}} \right)$
Simplifying the LHS term we get
$\Rightarrow \cos \left( {225} \right) = 1 - 2{\sin ^2}\left( {112\dfrac{1}{2}} \right)$
$\Rightarrow \cos \left( {180 + 45} \right) = 1 - 2{\sin ^2}\left( {112\dfrac{1}{2}} \right)$
$\Rightarrow \cos \left( {45} \right) = 1 - 2{\sin ^2}\left( {112\dfrac{1}{2}} \right)$
This cosine will lie in the third quadrant and the cosine trigonometry ratio in the third quadrant is negative.
The table of cosine function for standard angles is given as

Angle	0	30	45	60	90
cos	1	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{2}$	$0$

Therefore the above equation is written as
$\Rightarrow - \dfrac{1}{{\sqrt 2 }} = 1 - 2{\sin ^2}\left( {112\dfrac{1}{2}} \right)$
Take $- \dfrac{1}{{\sqrt 2 }}$ to RHS and $2{\sin ^2}\left( {112\dfrac{1}{2}} \right)$ to LHS, so the equation is written as
$\Rightarrow 2{\sin ^2}\left( {112\dfrac{1}{2}} \right) = 1 + \dfrac{1}{{\sqrt 2 }}$
On simplifying the RHS we get
$\Rightarrow 2{\sin ^2}\left( {112\dfrac{1}{2}} \right) = \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 }}$
Multiply and divide by the $\sqrt 2$we get
$\Rightarrow 2{\sin ^2}\left( {112\dfrac{1}{2}} \right) = \dfrac{{2 + \sqrt 2 }}{2}$
Divide the above equation by 2 we get
$\Rightarrow {\sin ^2}\left( {112\dfrac{1}{2}} \right) = \dfrac{{2 + \sqrt 2 }}{4}$
Take the square root on both sides. In LHS the square and the square root will get canceled. So we get
$\Rightarrow \sin \left( {112\dfrac{1}{2}} \right) = \pm \sqrt {\dfrac{{2 + \sqrt 2 }}{4}}$
The square root is implied to both the numerator and denominator and we get
$\Rightarrow \sin \left( {112\dfrac{1}{2}} \right) = \pm \dfrac{{\sqrt {2 + \sqrt 2 } }}{2}$
Therefore the value of $\sin \left( {112\dfrac{1}{2}} \right)$ using the double angle formula is $\pm \dfrac{{\sqrt {2 + \sqrt 2 } }}{2}$. If we use a half angle formula then the result will be the same.
Hence we have determined the value.
So, the correct answer is “ $\pm \dfrac{{\sqrt {2 + \sqrt 2 } }}{2}$”.

Note : In the question they have already mentioned to solve the given problem using the double or half angle formula. Therefore we must know about the formula. Here we have used the double angle formula $\cos 2\theta = 1 - 2{\sin ^2}\theta$, with the help of a table of trigonometry ratios for standard angle we have determined the value.