Question

If $\alpha =\dfrac{\pi }{3}$ , prove that $\cos \alpha \cos 2\alpha \cos 3\alpha \cos 4\alpha \cos 5\alpha \cos 6\alpha =-\dfrac{1}{16}$ .

Answer 1

Hint: Try to simplify the left-hand side of the equation that we need to prove by using the property that $\cos \left( 90{}^\circ -\alpha \right)=\sin \alpha $ , $\cos \left( 90{}^\circ +\alpha \right)=-\sin \alpha $ , and other similar formulas.

Complete step-by-step answer:
Before moving to the solution, let us discuss the periodicity of sine and cosine function, which we would be using in the solution. All the trigonometric ratios, including sine and cosine, are periodic functions. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.

Next, let us see the graph of cosx.

Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. $2{{\pi }^{c}}$ . So, we can say that the fundamental period of the cosine function and the sine function is $2{{\pi }^{c}}=360{}^\circ $
We will now solve the left-hand side of the equation given in the question.
$\cos \alpha \cos 2\alpha \cos 3\alpha \cos 4\alpha \cos 5\alpha \cos 6\alpha $
Putting the value $\alpha =\dfrac{\pi }{3}$ , we get
$=\cos \dfrac{\pi }{3}\cos \dfrac{2\pi }{3}\cos \dfrac{3\pi }{3}\cos \dfrac{4\pi }{3}\cos \dfrac{5\pi }{3}\cos \dfrac{6\pi }{3}$
$=\cos \dfrac{\pi }{3}\cos \left( \pi -\dfrac{\pi }{3} \right)\cos \pi \cos \left( \pi +\dfrac{\pi }{3} \right)\cos \left( 2\pi -\dfrac{\pi }{3} \right)\cos 2\pi $
Now we know $\cos \left( \pi +x \right)=-\cos x\text{ and cos}\left( \pi -x \right)=-\cos x$ . On putting these values in our expression, we get
$\cos \dfrac{\pi }{3}\left( -\cos \dfrac{\pi }{3} \right)\left( -\cos 0 \right)\left( -\cos \dfrac{\pi }{3} \right)\cos \left( 2\pi -\dfrac{\pi }{3} \right)\cos 2\pi $
We also know $\cos \left( 2\pi -x \right)=\cos x$ .
$\cos \dfrac{\pi }{3}\left( -\cos \dfrac{\pi }{3} \right)\left( -\cos 0 \right)\left( -\cos \dfrac{\pi }{3} \right)\cos \dfrac{\pi }{3}\cos 0$
Now, the value of $\cos \dfrac{\pi }{3}=\dfrac{1}{2}\text{ and }\cos 0=1$ . Putting this in our expression, we get
$\dfrac{1}{2}\times \left( -\dfrac{1}{2} \right)\times \left( -1 \right)\times \left( -\dfrac{1}{2} \right)\times \dfrac{1}{2}\times 1$
$=-\dfrac{1}{16}$
As we have shown that the left-hand side of the equation given in the question is equal to the right-hand side of the equation in the question, which is equal to $\dfrac{-1}{16}$. Hence, we can say that we have proved the equation given in the question .


Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, you need to remember the properties related to complementary angles and trigonometric ratios. It is preferred that while dealing with questions as above, you must first try to observe the pattern of the consecutive terms before applying the formulas, as directly applying the formulas may complicate the question.