Question

 Find the value of cos 15°?

Answer 1

Hint: We know cos30° if somehow we can convert cos15° into cos30° then we can find the value of cos15°. From the trigonometric identities we know that $\cos 2\theta =2{{\cos }^{2}}\theta -1$, in this double angle formula if we put θ = 15° then we get cos15° in terms of cos30° hence we can find the value of cos15°.

Complete step by step answer:
In the given question, we have to find the value of cos15° but we know the value of cos30°. So, we are going to use the formula for cosine of double angle.
$\cos 2\theta =2{{\cos }^{2}}\theta -1$
There are other forms of cos2θ also like:
$\cos 2\theta =1-2{{\sin }^{2}}\theta $
$\cos 2\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }$
But we are using the form in which cos2θ is there because we need the value of cosine.
$\cos 2\theta =2{{\cos }^{2}}\theta -1$
Substituting the value of θ = 15° we get,
$\cos {{30}^{0}}=2{{\cos }^{2}}{{15}^{0}}-1$
We know that$\cos {{30}^{0}}=\dfrac{\sqrt{3}}{2}$. Substituting the value of cos30° in the above equation we get,
$\begin{align}
  & \dfrac{\sqrt{3}}{2}=2{{\cos }^{2}}{{15}^{0}}-1 \\
 & \Rightarrow \sqrt{3}=4{{\cos }^{2}}{{15}^{0}}-2 \\
 & \Rightarrow \dfrac{\sqrt{3}+2}{4}={{\cos }^{2}}{{15}^{0}} \\
 & \Rightarrow \cos {{15}^{0}}=\pm \dfrac{\sqrt{\sqrt{3}+2}}{2} \\
\end{align}$
And as 15° is angle in the first quadrant then cosine of angle in the first quadrant is always positive so rejecting the negative value of cos15°.
Hence, the value of cos15° is$\dfrac{\sqrt{\sqrt{3}+2}}{2}$.

Note: There is an alternative method of finding the value of cos15° by writing cos15° as$\cos \left( {{60}^{0}}-{{45}^{0}} \right)$ then use the formula of$\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$$\cos \left( {{15}^{0}} \right)=\cos \left( {{60}^{0}}-{{45}^{0}} \right)$
Using$\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$in the above formula in which A = 60° and B = 45°we get,
$\cos \left( {{60}^{0}}-{{45}^{0}} \right)=\cos {{60}^{0}}\cos {{45}^{0}}+\sin {{60}^{0}}\sin {{45}^{0}}$
Substituting the values of cos60°, sin60°, cos45° and sin45° in the above equation we get,
$\begin{align}
  & \cos {{15}^{0}}=\dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}} \\
 & \Rightarrow \cos {{15}^{0}}=\dfrac{\sqrt{3}+1}{2\sqrt{2}} \\
 & \\
\end{align}$
Rationalizing the above expression we get,
$\begin{align}
  & \cos {{15}^{0}}=\dfrac{\left( \sqrt{3}+1 \right)\sqrt{2}}{4} \\
 & \cos {{15}^{0}}=\dfrac{\sqrt{6}+\sqrt{2}}{4} \\
\end{align}$
You will be wondering if two different values of cos15° are obtained. Both the values are the same but the difference is only in way of writing.
We can show that $\dfrac{\sqrt{\sqrt{3}+2}}{2}=\dfrac{\sqrt{6}+\sqrt{2}}{4}$
In $\dfrac{\sqrt{\sqrt{3}+2}}{2}$multiply 2 in both numerator and denominator then you will get,
$\begin{align}
  & \dfrac{\sqrt{4\sqrt{3}+8}}{4} \\
 & \dfrac{\sqrt{2.\sqrt{2}.\sqrt{2}.\sqrt{3}+2+6}}{4} \\
 & \dfrac{\sqrt{2.\sqrt{2}.\sqrt{6}+{{\left( \sqrt{2} \right)}^{2}}+{{\left( \sqrt{6} \right)}^{2}}}}{4} \\
 & \dfrac{\sqrt{{{\left( \sqrt{2}+\sqrt{6} \right)}^{2}}}}{4} \\
 & \dfrac{\sqrt{2}+\sqrt{6}}{4} \\
\end{align}$
Hence, we have shown that:
$\dfrac{\sqrt{\sqrt{3}+2}}{2}=\dfrac{\sqrt{6}+\sqrt{2}}{4}$