Question 41 of 50advanced🔧 ApplyLong Answer4 marks

Prove that:

2cos⁡π13cos⁡9π13+cos⁡3π13+cos⁡5π13=02 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0
Correct Answer

L.H.S. =

2cos⁡π13cos⁡9π13+cos⁡3π13+cos⁡5π132\cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}

Using the identity 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A \cos B = \cos(A+B) + \cos(A-B) on the first term:

=cos⁡(π13+9π13)+cos⁡(π13−9π13)+cos⁡3π13+cos⁡5π13= \cos\left(\frac{\pi}{13} + \frac{9\pi}{13}\right) + \cos\left(\frac{\pi}{13} - \frac{9\pi}{13}\right) + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} =cos⁡10π13+cos⁡(−8π13)+cos⁡3π13+cos⁡5π13= \cos \frac{10\pi}{13} + \cos \left(-\frac{8\pi}{13}\right) + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}

Since cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos \theta:

=cos⁡10π13+cos⁡8π13+cos⁡3π13+cos⁡5π13= \cos \frac{10\pi}{13} + \cos \frac{8\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}

Now, 10π13=π−3π13\frac{10\pi}{13} = \pi - \frac{3\pi}{13} and 8π13=π−5π13\frac{8\pi}{13} = \pi - \frac{5\pi}{13}

Therefore,

cos⁡10π13=cos⁡(π−3π13)=−cos⁡3π13\cos \frac{10\pi}{13} = \cos\left(\pi - \frac{3\pi}{13}\right) = -\cos \frac{3\pi}{13}

and

cos⁡8π13=cos⁡(π−5π13)=−cos⁡5π13\cos \frac{8\pi}{13} = \cos\left(\pi - \frac{5\pi}{13}\right) = -\cos \frac{5\pi}{13}

Substituting these values:

=−cos⁡3π13−cos⁡5π13+cos⁡3π13+cos⁡5π13= -\cos \frac{3\pi}{13} - \cos \frac{5\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} =0= 0

= R.H.S.

Hence proved.

Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 3 | Q: 1 | (Page 29)
For More Understanding

Explanation

This question from the Miscellaneous Exercise tests the application of the product-to-sum identity 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A \cos B = \cos(A+B) + \cos(A-B), which is demonstrated in Example 22 of the textbook.

The key insight is recognizing that angles like 10π13\frac{10\pi}{13} and 8π13\frac{8\pi}{13} can be expressed as π\pi minus smaller angles, allowing simplification using cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos \theta.

This technique of expressing larger angles in terms of complementary angles within the π\pi range is a standard approach in trigonometric proofs.

Solution Steps

  1. Step 1: Apply the identity 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A \cos B = \cos(A+B) + \cos(A-B) to transform the first term

  2. Step 2: Simplify using cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos \theta

  3. Step 3: Express 10π13\frac{10\pi}{13} and 8π13\frac{8\pi}{13} as π−3π13\pi - \frac{3\pi}{13} and π−5π13\pi - \frac{5\pi}{13} respectively

  4. Step 4: Apply cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos \theta to get negative terms

  5. Step 5: Combine like terms to get 00

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