Question

How do you find the exact value of $\cos 36$ using the sum and difference, double angle or half angle formulas?

Answer 1

Hint: We first take $\theta =36,A=18$ for the equation $2A=90-3A$. We take sine to use the formulas $\sin 2A=2\sin A\cos A$, $\cos 3A=4{{\cos }^{3}}A-3\cos A$, \[\cos 2\beta =1-2{{\sin }^{2}}\beta \]. We get the value of \[\sin 18=\dfrac{-1+\sqrt{5}}{4}\] which gives the value of $\cos 36$.

Complete step-by-step answer:
We need to find the value of $\cos 36$ using the sum and difference, double angle or half angle formulas.
We first assume $\theta =36,A=18$. Therefore, $5A=5\times 18=90$.
The equation gives $5A=90\Rightarrow 2A=90-3A$.
Now we take the trigonometric ratio sine on both sides $\sin 2A=\sin \left( 90-3A \right)$.
We know the formula of $\sin \left( 90-\alpha \right)=\cos \alpha $.
This gives $\sin 2A=\sin \left( 90-3A \right)=\cos 3A$.
Now we use the formulas of multiple angles where $\sin 2A=2\sin A\cos A$ and $\cos 3A=4{{\cos }^{3}}A-3\cos A$.
We put the values and get
\[\begin{align}
  & \sin 2A=\cos 3A \\
 & \Rightarrow 2\sin A\cos A=4{{\cos }^{3}}A-3\cos A \\
\end{align}\]
We take all the terms on one side and take $\cos A$ common and get
\[\begin{align}
  & 2\sin A\cos A-4{{\cos }^{3}}A+3\cos A=0 \\
 & \Rightarrow \cos A\left( 2\sin A-4{{\cos }^{2}}A+3 \right)=0 \\
\end{align}\]
We get multiplication of two terms as 0 which gives at least one of them has to be 0.
Now we know that $\cos A=\cos 18\ne 0$.
This gives \[\left( 2\sin A-4{{\cos }^{2}}A+3 \right)=0\].
We convert the term \[{{\cos }^{2}}A\] into \[{{\sin }^{2}}A\] using \[{{\cos }^{2}}A=1-{{\sin }^{2}}A\].
\[\begin{align}
  & 2\sin A-4{{\cos }^{2}}A+3=0 \\
 & \Rightarrow 2\sin A-4\left( 1-{{\sin }^{2}}A \right)+3=0 \\
 & \Rightarrow 2\sin A+4{{\sin }^{2}}A-1=0 \\
\end{align}\]
This is quadratic equation of $\sin A$ where we assume $x=\sin A$. We get \[4{{x}^{2}}+2x-1=0\].
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of $x$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have \[4{{x}^{2}}+2x-1=0\]. The values of a, b, c are $4,2,-1$ respectively.
We put the values and get $x$ as \[x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\times \left( -1 \right)\times 4}}{2\times 4}=\dfrac{-2\pm \sqrt{20}}{8}=\dfrac{-1\pm \sqrt{5}}{4}\].
Therefore, \[x=\sin A=\sin 18=\dfrac{-1+\sqrt{5}}{4}\]. The value \[x=\dfrac{-1-\sqrt{5}}{4}\] is not considered as the value of $\sin 18>0$.
Now we use \[\cos 2\beta =1-2{{\sin }^{2}}\beta \]. Putting the value of
We get $\cos 36=1-2{{\sin }^{2}}18=1-2{{\left( \dfrac{-1-\sqrt{5}}{4} \right)}^{2}}=1-\dfrac{3-\sqrt{5}}{4}=\dfrac{1+\sqrt{5}}{4}$.
Therefore, the value of $\cos 36$ is $\dfrac{1+\sqrt{5}}{4}$.

Note: Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty $. In that case we have to use the formula $x=n\pi \pm a$ for $\cos \left( x \right)=\cos a$ where $0\le a\le \pi $.