How do you find the exact value of $ \cos {15^ \circ } $ using the half angle formula?
Answer
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Hint: To find the exact value of $ \cos {15^ \circ } $ using the half angle formula. You are required to use trigonometric identities. In this question we use trigonometric identity of $ \cos 2\theta $ and modify it to find the required value of $ \cos {15^ \circ } $ . Also use the fact that $ {30^ \circ } $ can be written as $ {30^ \circ } = 2({15^ \circ }) $ .In this type of question you have to remember these trigonometric identities.
Complete step-by-step answer:
To solve this question we will first get half angle formula of cosine function from
$ \cos 2\theta $ and after which we will put the value $ {15^ \circ } $ in the formula to get the exact value of $ \cos {15^ \circ } $ using half angle formula.
So here is the formula of $ \cos 2\theta $
$ \cos 2\theta = 2{\cos ^2}\theta - 1 $
And we that $ {30^ \circ } $ can be written as $ {30^ \circ } = 2({15^ \circ }) $
So we will assume that $ 2\theta = {30^ \circ } $ which means that value of $ \theta = \dfrac{{{{30}^ \circ }}}{2} $
Implies that $ \theta = {15^ \circ } $
Putting value of $ \theta $ in the formula of $ \cos 2\theta $ we get,
$ \cos 2({15^ \circ }) = 2{\cos ^2}({15^ \circ }) - 1 $
Now taking $ 2{\cos ^2}({15^ \circ }) $ to LHS and $ \cos 2({30^ \circ }) $ to RHS we will get
$ 2{\cos ^2}({15^ \circ }) = \cos 2({15^ \circ }) + 1 $
$ 2{\cos ^2}({15^ \circ }) = \cos {30^ \circ } + 1 $
Putting value of $ \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2} $ , in the above equation we get
\[
2{\cos ^2}({15^ \circ }) = \dfrac{{\sqrt 3 }}{2} + 1 \\
2{\cos ^2}({15^ \circ }) = \dfrac{{\sqrt 3 + 2}}{2} \\
{\cos ^2}({15^ \circ }) = \dfrac{{\sqrt 3 + 2}}{4} \\
\;
\]
\[
\cos {15^ \circ } = \sqrt {\dfrac{{\sqrt 3 + 2}}{4}} \\
\cos {15^ \circ } = \pm \sqrt {\dfrac{{\sqrt 3 + 2}}{4}} \;
\]
Now we know that the value of $ \cos {15^ \circ } $ will be positive because $ {15^ \circ } $ lies in the first quadrant and the value of the cosine function in the first and fourth quadrant is positive. Hence value of $ \cos {15^ \circ } $ is equal to
$ \cos {15^ \circ } = \dfrac{{\sqrt {\sqrt 3 + 2} }}{2} $
So the exact value of $ \cos {15^ \circ } $ using the half angle formula is equal to $ \dfrac{{\sqrt {\sqrt 3 + 2} }}{2} $ .
So, the correct answer is “ $ \dfrac{{\sqrt {\sqrt 3 + 2} }}{2} $ ”.
Note: While solving this type of question you need to be careful about the sign of final value. For this you have to check in which quadrant trigonometric function value will be positive and in which it is negative. Knowing trigonometric formulas is a must to solve this kind of problem.
Complete step-by-step answer:
To solve this question we will first get half angle formula of cosine function from
$ \cos 2\theta $ and after which we will put the value $ {15^ \circ } $ in the formula to get the exact value of $ \cos {15^ \circ } $ using half angle formula.
So here is the formula of $ \cos 2\theta $
$ \cos 2\theta = 2{\cos ^2}\theta - 1 $
And we that $ {30^ \circ } $ can be written as $ {30^ \circ } = 2({15^ \circ }) $
So we will assume that $ 2\theta = {30^ \circ } $ which means that value of $ \theta = \dfrac{{{{30}^ \circ }}}{2} $
Implies that $ \theta = {15^ \circ } $
Putting value of $ \theta $ in the formula of $ \cos 2\theta $ we get,
$ \cos 2({15^ \circ }) = 2{\cos ^2}({15^ \circ }) - 1 $
Now taking $ 2{\cos ^2}({15^ \circ }) $ to LHS and $ \cos 2({30^ \circ }) $ to RHS we will get
$ 2{\cos ^2}({15^ \circ }) = \cos 2({15^ \circ }) + 1 $
$ 2{\cos ^2}({15^ \circ }) = \cos {30^ \circ } + 1 $
Putting value of $ \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2} $ , in the above equation we get
\[
2{\cos ^2}({15^ \circ }) = \dfrac{{\sqrt 3 }}{2} + 1 \\
2{\cos ^2}({15^ \circ }) = \dfrac{{\sqrt 3 + 2}}{2} \\
{\cos ^2}({15^ \circ }) = \dfrac{{\sqrt 3 + 2}}{4} \\
\;
\]
\[
\cos {15^ \circ } = \sqrt {\dfrac{{\sqrt 3 + 2}}{4}} \\
\cos {15^ \circ } = \pm \sqrt {\dfrac{{\sqrt 3 + 2}}{4}} \;
\]
Now we know that the value of $ \cos {15^ \circ } $ will be positive because $ {15^ \circ } $ lies in the first quadrant and the value of the cosine function in the first and fourth quadrant is positive. Hence value of $ \cos {15^ \circ } $ is equal to
$ \cos {15^ \circ } = \dfrac{{\sqrt {\sqrt 3 + 2} }}{2} $
So the exact value of $ \cos {15^ \circ } $ using the half angle formula is equal to $ \dfrac{{\sqrt {\sqrt 3 + 2} }}{2} $ .
So, the correct answer is “ $ \dfrac{{\sqrt {\sqrt 3 + 2} }}{2} $ ”.
Note: While solving this type of question you need to be careful about the sign of final value. For this you have to check in which quadrant trigonometric function value will be positive and in which it is negative. Knowing trigonometric formulas is a must to solve this kind of problem.
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