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Question: If \(\alpha =\dfrac{\pi }{3}\) , prove that \(\cos \alpha \cos 2\alpha \cos 3\alpha \cos 4\alpha \co...

If α=π3\alpha =\dfrac{\pi }{3} , prove that cos⁡αcos⁡2αcos⁡3αcos⁡4αcos⁡5αcos⁡6α=−116\cos \alpha \cos 2\alpha \cos 3\alpha \cos 4\alpha \cos 5\alpha \cos 6\alpha =-\dfrac{1}{16} .

Explanation

Solution

Hint: Try to simplify the left-hand side of the equation that we need to prove by using the property that cos⁡(90∘−α)=sin⁡α\cos \left( 90{}^\circ -\alpha \right)=\sin \alpha , cos⁡(90∘+α)=−sin⁡α\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. 2πc2{{\pi }^{c}} . So, we can say that the fundamental period of the cosine function and the sine function is 2πc=360∘2{{\pi }^{c}}=360{}^\circ
We will now solve the left-hand side of the equation given in the question.
cos⁡αcos⁡2αcos⁡3αcos⁡4αcos⁡5αcos⁡6α\cos \alpha \cos 2\alpha \cos 3\alpha \cos 4\alpha \cos 5\alpha \cos 6\alpha
Putting the value α=π3\alpha =\dfrac{\pi }{3} , we get
=cos⁡π3cos⁡2π3cos⁡3π3cos⁡4π3cos⁡5π3cos⁡6π3=\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⁡π3cos⁡(π−π3)cos⁡πcos⁡(π+π3)cos⁡(2π−π3)cos⁡2π=\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⁡(π+x)=−cos⁡x and cos(π−x)=−cos⁡x\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⁡π3(−cos⁡π3)(−cos⁡0)(−cos⁡π3)cos⁡(2π−π3)cos⁡2π\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⁡(2π−x)=cos⁡x\cos \left( 2\pi -x \right)=\cos x .
cos⁡π3(−cos⁡π3)(−cos⁡0)(−cos⁡π3)cos⁡π3cos⁡0\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⁡π3=12 and cos⁡0=1\cos \dfrac{\pi }{3}=\dfrac{1}{2}\text{ and }\cos 0=1 . Putting this in our expression, we get
12×(−12)×(−1)×(−12)×12×1\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
=−116=-\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 −116\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.