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:
$\cos 2\theta =1-2{\sin }^2\theta$
$\cos 2\theta ={\displaystyle \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }}$
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$\cos {30}^0={\displaystyle \frac{\sqrt{3}}{2}}$. Substituting the value of cos30° in the above equation we get,
$\begin{array}{rl}& {\displaystyle \frac{\sqrt{3}}{2}}=2{\cos }^2{15}^0-1\\ & \Rightarrow \sqrt{3}=4{\cos }^2{15}^0-2\\ & \Rightarrow {\displaystyle \frac{\sqrt{3}+2}{4}}={\cos }^2{15}^0\\ & \Rightarrow \cos {15}^0=\pm {\displaystyle \frac{\sqrt{\sqrt{3}+2}}{2}}\end{array}$
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${\displaystyle \frac{\sqrt{\sqrt{3}+2}}{2}}$.

Note: There is an alternative method of finding the value of cos15° by writing cos15° as$\cos ({60}^0-{45}^0)$ then use the formula of$\cos (A-B)=\cos A\cos B+\sin A\sin B$$\cos \left({15}^0\right)=\cos ({60}^0-{45}^0)$
Using$\cos (A-B)=\cos A\cos B+\sin A\sin B$in the above formula in which A = 60° and B = 45°we get,
$\cos ({60}^0-{45}^0)=\cos {60}^0\cos {45}^0+\sin {60}^0\sin {45}^0$
Substituting the values of cos60°, sin60°, cos45° and sin45° in the above equation we get,
$\begin{array}{rl}& \cos {15}^0={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{\sqrt{3}}{2}}\times {\displaystyle \frac{1}{\sqrt{2}}}\\ & \Rightarrow \cos {15}^0={\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}}\\ & \end{array}$
Rationalizing the above expression we get,
$\begin{array}{rl}& \cos {15}^0={\displaystyle \frac{(\sqrt{3}+1)\sqrt{2}}{4}}\\ & \cos {15}^0={\displaystyle \frac{\sqrt{6}+\sqrt{2}}{4}}\end{array}$
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 ${\displaystyle \frac{\sqrt{\sqrt{3}+2}}{2}}={\displaystyle \frac{\sqrt{6}+\sqrt{2}}{4}}$
In ${\displaystyle \frac{\sqrt{\sqrt{3}+2}}{2}}$multiply 2 in both numerator and denominator then you will get,
$\begin{array}{rl}& {\displaystyle \frac{\sqrt{4\sqrt{3}+8}}{4}}\\ & {\displaystyle \frac{\sqrt{2.\sqrt{2}.\sqrt{2}.\sqrt{3}+2+6}}{4}}\\ & {\displaystyle \frac{\sqrt{2.\sqrt{2}.\sqrt{6}+{\left(\sqrt{2}\right)}^2+{\left(\sqrt{6}\right)}^2}}{4}}\\ & {\displaystyle \frac{\sqrt{{(\sqrt{2}+\sqrt{6})}^2}}{4}}\\ & {\displaystyle \frac{\sqrt{2}+\sqrt{6}}{4}}\end{array}$
Hence, we have shown that:
${\displaystyle \frac{\sqrt{\sqrt{3}+2}}{2}}={\displaystyle \frac{\sqrt{6}+\sqrt{2}}{4}}$