CosA + cosB + cosC = 1 + 4sinA/2 sinB/2 sinC/2

1 answer:

6 0

$\cos(A) + \cos( B ) + \cos(C) \\ = 2 \cos( \frac{A + B}{2} ) \cos( \frac{ A- B}{2} ) + 1 - 2 { \sin }^{2} \frac{C}{2} \\ = 1 + 2\cos( \frac{\pi - C}{2} ) \cos( \frac{A - B}{2} ) - 2 { \sin }^{2} \frac{C}{2} \\ = 1 + 2 \sin \frac{C}{2} \cos( \frac{A - B}{2} ) - 2 { \sin }^{2} \frac{C}{2} \\ = 1 + 2 \sin \frac{C}{2} [ \cos( \frac{A - B}{2} ) - \sin \frac{C}{2}] \\ = 1 + 2 \sin \frac{C}{2} [ \cos( \frac{A - B}{2} ) - \cos( \frac{A + B}{2} )] \\ = 1 + 2 \sin\frac{C}{2} \times 2 \sin( \frac{ \frac{A + B}{2} +\frac{A - B}{2} }{2} ) \sin( \frac{ \frac{A + B}{2} - \frac{A - B}{2} }{2} ) \\ = 1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

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The standard form of the quadratic equation is x² + 3·x - 4 = 0

Step-by-step explanation:

The standard form of a quadratic equation is a·x² + b·x + c = 0

Given that the expression of the quadratic equation is (x + 4)·(x - 1) = y, we can write the given expression in standard form by expanding, and equating the result to zero as follows;

(x + 4)·(x - 1) = x² - x + 4·x - 4 = x² + 3·x - 4 = 0

The standard form of the quadratic equation is x² + 3·x - 4 = 0

The graph of the equation created with MS Excel is attached