Question

How do you prove $\cos 36\cdot \cos 72=\dfrac{1}{4}$?

Answer 1

Hint: For proving the given expression we have to use the identity $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$ on the LHS. Then, with the help of the trigonometric identities \[\cos \left( -x \right)=\cos x\] and $\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta $ we will obtain the LHS as a difference of two cosines. Finally, on using the trigonometric identity $\cos 2A=2{{\cos }^{2}}A-1$ and the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we will be able to prove the LHS to be equal to the RHS.

Complete step by step answer:
The expression to be proved is given in the question as
$\cos {{36}^{\circ }}\cdot \cos {{72}^{\circ }}=\dfrac{1}{4}$
Considering the LHS of the above equation, we have
$\Rightarrow LHS=\cos {{36}^{\circ }}\cdot \cos {{72}^{\circ }}$
Multiplying and dividing the above expression by $2$ we get
$\Rightarrow LHS=\dfrac{2\cos {{36}^{\circ }}\cdot \cos {{72}^{\circ }}}{2}.......(i)$
Now, we know the trigonometric identity
$2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$
Substituting \[A={{36}^{\circ }}\] and \[B={{72}^{\circ }}\] in the above identity, we get
$\begin{align}
  & \Rightarrow 2\cos {{36}^{\circ }}\cdot \cos {{72}^{\circ }}=\cos \left( {{36}^{\circ }}+{{72}^{\circ }} \right)+\cos \left( {{36}^{\circ }}-{{72}^{\circ }} \right) \\
 & \Rightarrow 2\cos {{36}^{\circ }}\cdot \cos {{72}^{\circ }}=\cos \left( {{108}^{\circ }} \right)+\cos \left( -{{36}^{\circ }} \right) \\
\end{align}$
Now, we know that \[\cos \left( -x \right)=\cos x\]. So the above equation can be written as
$\Rightarrow 2\cos {{36}^{\circ }}\cdot \cos {{72}^{\circ }}=\cos \left( {{108}^{\circ }} \right)+\cos \left( {{36}^{\circ }} \right)$
Substituting this in the equation (i) we get
\[\Rightarrow LHS=\dfrac{1}{2}\left( \cos \left( {{108}^{\circ }} \right)+\cos \left( {{36}^{\circ }} \right) \right)\]
Now, writing the angle \[{{108}^{\circ }}={{180}^{\circ }}-{{72}^{\circ }}\] we get
\[\Rightarrow LHS=\dfrac{1}{2}\left( \cos \left( {{180}^{\circ }}-{{72}^{\circ }} \right)+\cos \left( {{36}^{\circ }} \right) \right)\]
Now, we know that $\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta $. So the above equation can be written as
\[\begin{align}
  & \Rightarrow LHS=\dfrac{1}{2}\left( -\cos \left( {{72}^{\circ }} \right)+\cos \left( {{36}^{\circ }} \right) \right) \\
 & \Rightarrow LHS=\dfrac{1}{2}\left( \cos {{36}^{\circ }}-\cos {{72}^{\circ }} \right) \\
\end{align}\]
Now, let $x=\cos {{36}^{\circ }}$ and $y=\cos {{72}^{\circ }}$. Putting these above, we get
\[\Rightarrow LHS=\dfrac{1}{2}\left( x-y \right)...........(ii)\]
We know that $\cos 2A=2{{\cos }^{2}}A-1$.
Substituting $A={{36}^{\circ }}$ in the above identity, we get
\[\begin{align}
  & \Rightarrow \cos 2\left( {{36}^{\circ }} \right)=2{{\cos }^{2}}{{36}^{\circ }}-1 \\
 & \Rightarrow \cos {{72}^{\circ }}=2{{\cos }^{2}}{{36}^{\circ }}-1 \\
 & \Rightarrow y=2{{x}^{2}}-1...........(iii) \\
\end{align}\]
Similarly, substituting $A={{72}^{\circ }}$ in the above identity, we get
\[\begin{align}
  & \Rightarrow \cos 2\left( {{72}^{\circ }} \right)=2{{\cos }^{2}}{{72}^{\circ }}-1 \\
 & \Rightarrow \cos {{144}^{\circ }}=2{{\cos }^{2}}{{72}^{\circ }}-1 \\
 & \Rightarrow \cos {{144}^{\circ }}=2{{y}^{2}}-1 \\
\end{align}\]
Now, writing the angle \[{{144}^{\circ }}={{180}^{\circ }}-{{36}^{\circ }}\] in the above equation, we get
\[\Rightarrow \cos \left( {{180}^{\circ }}-{{36}^{\circ }} \right)=2{{y}^{2}}-1\]
Now, from the identity $\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta $ we can write the above equation as
\[\begin{align}
  & \Rightarrow -\cos {{36}^{\circ }}=2{{y}^{2}}-1 \\
 & \Rightarrow -x=2{{y}^{2}}-1...........(iv) \\
\end{align}\]
Subtracting (iv) from (iii) we get
$\begin{align}
  & \Rightarrow y-\left( -x \right)=2{{x}^{2}}-1-\left( 2{{y}^{2}}-1 \right) \\
 & \Rightarrow y+x=2{{x}^{2}}-2{{y}^{2}}-1+1 \\
 & \Rightarrow y-\left( -x \right)=2{{x}^{2}}-1-\left( 2{{y}^{2}}-1 \right) \\
 & \Rightarrow y+x=2\left( {{x}^{2}}-{{y}^{2}} \right) \\
\end{align}$
Now, from the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we write the above equation as
$\begin{align}
  & \Rightarrow \left( y+x \right)=2\left( x+y \right)\left( x-y \right) \\
 & \Rightarrow \left( x+y \right)=2\left( x+y \right)\left( x-y \right) \\
\end{align}$
Subtracting $\left( x+y \right)$ from both the sides, we get
\[\begin{align}
  & \Rightarrow \left( x+y \right)-\left( x+y \right)=2\left( x+y \right)\left( x-y \right)-\left( x+y \right) \\
 & \Rightarrow 0=\left( x+y \right)\left( 2\left( x-y \right)-1 \right) \\
 & \Rightarrow \left( x+y \right)\left( 2\left( x-y \right)-1 \right)=0 \\
\end{align}\]
On solving we get
$\left( x+y \right)=0$ and $2\left( x-y \right)-1=0$
We have $x=\cos {{36}^{\circ }}$ and $y=\cos {{72}^{\circ }}$. Since both the angles are less than ${{90}^{\circ }}$, so $\left( x+y \right)\ne 0$. Thus, we have
$\begin{align}
  & \Rightarrow 2\left( x-y \right)-1=0 \\
 & \Rightarrow \left( x-y \right)=\dfrac{1}{2} \\
\end{align}$
Substituting this in (ii), we get
$\begin{align}
  & \Rightarrow LHS=\dfrac{1}{2}\left( \dfrac{1}{2} \right) \\
 & \Rightarrow LHS=\dfrac{1}{4} \\
 & \Rightarrow LHS=RHS \\
\end{align}$

Hence, the given expression $\cos 36\cdot \cos 72=\dfrac{1}{4}$ is proved.

Note: For solving this question, we need to have a fair idea regarding the trigonometric identities used in the above solution. Also, we must not be confused regarding the negative sign appearing in the identity $\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta $. The angle ${{180}^{\circ }}-\theta $ belongs to the second quadrant, where the cosine function is negative. This justifies the negative sign present in the identity.