Question

Find the values of $\cos \left(72°\right)$.

Answer 1

Step 1: Find the value of $\sin \left(18°\right)$

Let $A=18°$

$$\begin{gathered} \Rightarrow 5A=90° \\ \Rightarrow 2A+3A=90° \\ \Rightarrow 2A=90°-3A \\ \Rightarrow \sin \left(2A\right)=\sin \left(90°-3A\right) \\ \Rightarrow \sin \left(2A\right)=\cos \left(3A\right);\left[\because \sin (90-\theta )=\cos \theta \right] \\ \Rightarrow 2\sin \left(A\right)\cos \left(A\right)=4{\cos }^3\left(A\right)-3\cos \left(A\right);\left[\cos 3A=4{\cos }^{3}A-3\cos A\right] \\ \Rightarrow 4{\cos }^3\left(A\right)-3\cos \left(A\right)-2\sin \left(A\right)\cos \left(A\right)=0 \\ \Rightarrow \cos \left(A\right)\left(4{\cos }^2\left(A\right)-2\sin \left(A\right)-3\right)=0 \end{gathered}$$

Now,

$$\begin{gathered} \cos \left(A\right)\ne 0 \\ \Rightarrow 4{\cos }^2\left(A\right)-2\sin \left(A\right)-3=0 \\ \Rightarrow 4\left(1-{\sin }^2\left(A\right)\right)-2\sin \left(A\right)-3=0 \\ \Rightarrow 4{\sin }^2\left(A\right)+2\sin \left(A\right)-1=0 \\ \Rightarrow \sin \left(A\right)=\frac{-2\pm \sqrt{4-4\left(4\right)\left(-1\right)}}{2\times 4}\left[\because x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right] \\ =\frac{-1\pm \sqrt{5}}{4} \end{gathered}$$

$\sin \left(A\right)$ is positive as$18°$lies in the first quadrant. Therefore,

$\Rightarrow \sin \left(A\right)=\frac{\sqrt{5}-1}{4}$

Step 2: Use $\sin \left(18°\right)$to find $\cos \left(72°\right)$

Now,

$$\begin{gathered} \Rightarrow \cos \left(72°\right)=\cos \left(90°-18°\right)=\sin \left(18°\right)\left[\because \cos (90-\theta )=\sin \theta \right] \\ =\frac{\sqrt{5}-1}{4} \end{gathered}$$

Hence, the value of $\cos \left(72°\right)$ is $\frac{\sqrt{5}-1}{4}$