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Principal Solution of Trigonometric Equation

Prove that in...

Question

Prove that in any $△$ ABC, $c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ , where a, b and c are the magnitudes of the sides opposite to the vertices A, B and C, respectively.

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Solution

Here, components of C are c cos A and c sin A is drawn.

$Since, - - \to C D = b - c c o s A I n △ B D C, a^{2} = (b - c c o s A)^{2} + (c s i n A)^{2} \Rightarrow a^{2} = b^{2} + c^{2} c o s^{2} A - 2 b c c o s A + c^{2} s i n^{2} A \Rightarrow 2 b c c o s A = b^{2} - a^{2} + c^{2} (c o s^{2} A + s i n^{2} A) ∴ c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$

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