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Question: How do you use the half-angle formula to find the exact values of \(\sin {22.5^ \circ }\)?...

How do you use the half-angle formula to find the exact values of sin22.5\sin {22.5^ \circ }?

Explanation

Solution

For solving this question we will use the identity cos2θ=1sin2θ\cos 2\theta = 1 - {\sin ^2}\theta . Then, make this identity equal to sin2θ{\sin ^2}\theta and put 22.5{22.5^ \circ } in the place of θ\theta in the whole identity. Then, put the value of cos45\cos {45^ \circ } equal to 12\dfrac{1}{{\sqrt 2 }}.

Complete Step by Step Solution:
Half-angle identities are a set of equations that help you translate the trigonometric values of unfamiliar angles into more familiar values. There are many half-angle identities but we use them according to the necessity of the question.
We have to find the value of sin22.5\sin {22.5^ \circ } where angle is 22.5{22.5^ \circ }. As already stated in the question we have to use the half angle formula, so, we will use the formula of cos2θ\cos 2\theta . So, we know that –
cos2θ=1sin2θ\cos 2\theta = 1 - {\sin ^2}\theta
By transposition method making this formula equal to sin2θ{\sin ^2}\theta , we get –
sin2θ=12[1cos2θ]\Rightarrow {\sin ^2}\theta = \dfrac{1}{2}\left[ {1 - \cos 2\theta } \right]
We have to find the value of sin22.5\sin {22.5^ \circ } where the angle is 22.5{22.5^ \circ }. Putting this value of angle in the place of θ\theta in the above identity, we get –
sin222.5=12[1cos2×22.5] sin222.5=12[1cos45]  \Rightarrow {\sin ^2}22.5 = \dfrac{1}{2}\left[ {1 - \cos 2 \times 22.5} \right] \\\ \Rightarrow {\sin ^2}22.5 = \dfrac{1}{2}\left[ {1 - \cos 45} \right] \\\
Putting the value of cos45\cos {45^ \circ } on the above equation, we get –
sin222.5=12[112]\Rightarrow {\sin ^2}22.5 = \dfrac{1}{2}\left[ {1 - \dfrac{1}{{\sqrt 2 }}} \right]
Now, further solving, we get –
sin222.5=12[212] sin222.5=2122  \Rightarrow {\sin ^2}22.5 = \dfrac{1}{2}\left[ {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}} \right] \\\ \Rightarrow {\sin ^2}22.5 = \dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }} \\\
Now, doing the square root on the both sides, we get –
sin22.5=±2122\Rightarrow \sin 22.5 = \pm \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}}
But we will discard the negative value of the sin22.5\sin 22.5 and considering only positive value, we get –
sin22.5=2122\sin 22.5 = \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}}

Hence, the value of sin22.5\sin 22.5 is 2122\sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}}

So, this is the required value of sin22.5\sin 22.5.

Note: Use the half-angle formula only which has sinθ\sin \theta to solve this question. Many students can make mistakes while taking the value of sin22.5\sin 22.5 only positive and not negative as the value of sin\sin is only positive from the angle 00 to 90{90^ \circ }.