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never [62]
3 years ago
6

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

Mathematics
1 answer:
bulgar [2K]3 years ago
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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