In a triangle ABC- cosA-cos B-A- -2 sin C - 2 sin A - B - 2B- -2 sin C - 2 sin A - B - 2C- -2 cos C - 2 sin A-B - 2D- 2 cos C - 2 sin A - B - 2

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Calculating Heights and Distances

In a triangle...

Question

In a triangle ABC,
cosA- cosB=

-2sin(C/2) sin(A+B/2)

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- 2sin(C/2) sin(A-B/2)

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-2cos(C/2) sin(A-B)/2

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2cos(C/2) sin(A+B/2)

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Solution

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Similar questions

A, B

C

cos A + cos B + cos C - 1

2 sin \frac{A}{2} c o s e c \frac{B}{2} sin \frac{C}{2} - sin A cot \frac{B}{2} - cos A

A + B + C = π

sin A + sin B - sin C = 2 sin \frac{A}{2} sin \frac{B}{2} cos \frac{C}{2}

Δ

A B C

a, b, c

A . P

\frac{2 S i n \frac{A}{2} S i n \frac{C}{2}}{S i n \frac{B}{2}} =

∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} s i n \frac{A}{2} & s i n \frac{B}{2} & s i n \frac{c}{2} s i n (A + B + C) & s i n \frac{B}{2} & c o s \frac{A}{2} c o s (\frac{A + B + C}{2}) & t a n (A + B + C) & s i n \frac{C}{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

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