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# How do you find the exact value of $sin^2\left( {\pi/8} \right)$ using the half angle formula?

Last updated date: 26th Feb 2024
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Hint: For solving the expression given in the question,
Apply the formula of half angle,
As, ${\sin ^2}\left( x \right) = \dfrac{1}{2}\left[ {1 - \cos \left( {2x} \right)} \right]$
So, apply the above-mentioned formula to find the value of “x” , the value for angle that can be expressed in the form of half angle.

Complete step by step solution: As per data given in the question,
As we have to determine the value of angle of Sine,
For determining the value of $sin^2\left( {\pi/8} \right)$ ,
As here,
From the given expression, we can conclude that,
We have to solve the above element using the half angle formula.
First of all writing the half angle formula are as follow:
${\sin ^2}\left( x \right) = \dfrac{1}{2}\left[ {1 - \cos \left( {2x} \right)} \right]$
So,
Now for solving we have to consider
$x = \dfrac{\pi }{8}$
Then we can solve it as $2x = 2\dfrac{\pi }{8}$
That will be equal as
$2x = \dfrac{\pi }{4}$ i.e. $x = \dfrac{\pi }{8}$
So, solving above we get the value of $x$
Now substituting in the formula of half angle formula we get
${\sin ^2}\left( {\dfrac{\pi }{8}} \right) = \dfrac{1}{2}\left[ {1 - \cos \left( {\dfrac{\pi }{4}} \right)} \right]$
But here the value of
$\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}$
So, further we have to substitute the value of $\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}$ in above equation we get
$\sin \left( {\dfrac{\pi }{8}} \right) = \pm \sqrt {\dfrac{{2 - \sqrt 2 }}{4}}$
Thus the value we get
$= \pm 0.3827$
Thus now from the obtained value we have to choose the positive value. Hence the value is $0.3827$

As we know that,
Value of angle of sine and value of angle of cosecant are inverse of each other.
Value of angle of secant and value of angle of cosine are inverse of each other.
Value of angle of cotangent and value of angle of tangent are inverse to each other.
To the above formula we have used i.e. half angle identity as we can write as.
$\sin a = 2\sin \left( {\dfrac{a}{2}} \right).\cos \left( {\dfrac{a}{2}} \right)$
Therefore,
$t = \tan \dfrac{a}{2}$
Now further,
$\sin a = \dfrac{{2t}}{{1 + {t^2}}}$
$\cos a = \dfrac{{1 - {t^2}}}{{1 + {t^2}}}$
$\tan a = \dfrac{{2t}}{{1 - {t^2}}}$

Note:
We can also find the half angle for $\cos 15.$
Let $\cos \left( {{{15}^o}} \right) = \cos \left( {\dfrac{{30^\circ }}{2}} \right)$
Now, we can write half angle identity as
$\cos \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 + \cos \left( \theta \right)}}{2}}$
So further
$\cos \left( {15^\circ } \right) = \cos \left( {\dfrac{{30^\circ }}{2}} \right)$
$= \pm \sqrt {\dfrac{{1 + \cos \left( {30^\circ } \right)}}{2}}$
So, the value will be $\pm 0.966$