If $\cos3A + \cos3B + \cos3C = 1$ in a triangle, find one of its length

$\mathbf {Hint...}$

$$\cos {3A}+\cos {3B}+ \cos{3C}=1$$ $$\Rightarrow 4 \sin {\frac {3C}{2}} .\sin {\frac{3B}{2}} .\sin{\frac{3A}{2}}=0$$

Hence the largest angle of triangle is $\frac{2\pi}{3}$ which can be either angle $C$ or angle $A$. By applying cosine rule in each of these cases we get the value of $AB$ as $\sqrt {399}$ or $\sqrt {94}-5$ respectively.

Note:

$$\cos {3A}+\cos {3B}+ \cos{3C}=1$$ $$\Rightarrow -2\cos {\frac {3(A-B)}{2}}\sin {\frac {3C}{2}}- 2\left(\sin {\frac {3C}{2}}\right)^2=0$$

$$=\sin\frac{3C}{2}\left(\cos\frac{3(A-B)}{2}+ \sin\frac{3C}{2}\right)=0$$

$$\sin\frac{3C}{2}\left(\cos\frac{3(A-B)}{2}-\cos\frac{3(A+B)}{2}\right)=0$$

$$\sin\frac{3C}{2}\sin\frac{3B}{2}\sin\frac{3A}{2}=0.$$

Tags:

Trigonometry

Functions

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