Solutions of Triangle

Geometry Level 3

cos ⁡ ( A ) + cos ⁡ ( C ) = 4 sin ⁡ 2 ( B 2 ) \cos(A)+\cos(C)=4\sin^{2}\left(\frac{B}{2}\right) Then a , b , c a,b, c are in what kind of progression?

In triangle A B C ABC , a , b , c a,b,c are sides opposite to angle A , B , C A,B,C respectively.

Geometric Progression Harmonic Progression Arithmetic Progression None of these

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Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Given: cos ⁡ C + cos ⁡ B = 4 sin ⁡ 2 A 2 ⟹ 2 cos ⁡ ( B + C 2 ) cos ⁡ ( B − C 2 ) = 4 sin ⁡ 2 A 2 ⟹ cos ⁡ ( B − C 2 ) = 2 sin ⁡ A 2 ( ∵ cos ⁡ ( B + C 2 ) = cos ⁡ ( π − A 2 ) = sin ⁡ A 2 ≠ 0 ) ⟹ cos ⁡ ( B − C 2 ) × 2 sin ⁡ ( B + C 2 ) = 2 sin ⁡ A 2 × 2 cos ⁡ A 2 ( ∵ sin ⁡ ( B + C 2 ) = sin ⁡ ( π − A 2 ) = cos ⁡ A 2 ≠ 0 ) ⟹ sin ⁡ B + sin ⁡ C = 2 sin ⁡ A ⟹ b + c = 2 a \begin{aligned}&\textbf{Given: }\cos C+\cos B=4\sin^2\frac{A}{2}\\&\implies2\cos \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)=4\sin^2\frac{A}{2}\\&\implies\cos \left(\frac{B-C}{2}\right)=2\sin \frac{A}{2}\\&\left(\because \cos \left(\frac{B+C}{2}\right)=\cos \left(\frac{\pi-A}{2}\right)=\sin \frac{A}{2}\neq0\right)\\&\implies\cos \left(\frac{B-C}{2}\right)\times2\sin \left(\frac{B+C}{2}\right)=2\sin \frac{A}{2}\times2\cos \frac{A}{2}\\&\left(\because \sin \left(\frac{B+C}{2}\right)=\sin \left(\frac{\pi-A}{2}\right)=\cos \frac{A}{2}\neq0\right)\\&\implies\sin B+\sin C=2\sin A\\&\implies b+c=2a\end{aligned}

The last equation follows by an application of extended sine rule .

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We see that b,a,c are in AP and not a,b,c.Since the order matters,the question statement should be corrected.

Indraneel Mukhopadhyaya - 5 years, 2 months ago

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I agree with you.

FYI In future if you encounter any flaw/mistake in the problem please report it by clicking "rectangle-rectangle-rectangle" box in the top right corner of the problem, near level. Then select "view reports" to write a report accordingly. Thanks.

Nihar Mahajan - 5 years, 2 months ago

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