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Question Number 120393 by help last updated on 31/Oct/20

Commented by help last updated on 31/Oct/20

pls how to prove the identity?

Commented by $@y@m last updated on 31/Oct/20

Please type the question.

$${Please}\:{type}\:{the}\:{question}. \\ $$

Commented by help last updated on 31/Oct/20

sinA+sinB+sinC−(sinA+B+C)=??

$${sinA}+{sinB}+{sinC}−\left({sinA}+{B}+{C}\right)=?? \\ $$$$ \\ $$

Commented by $@y@m last updated on 31/Oct/20

=2sin ((A+B)/2)cos ((A−B)/2)+2cos ((A+B+2C)/2)sin ((−A−B)/2)  =2sin ((A+B)/2){cos ((A−B)/2)−cos ((A+B+2C)/2)}  =2sin ((A+B)/2).2sin  ((A+C)/2)sin ((B+C)/2)

$$=\mathrm{2sin}\:\frac{{A}+{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{2cos}\:\frac{{A}+{B}+\mathrm{2}{C}}{\mathrm{2}}\mathrm{sin}\:\frac{−{A}−{B}}{\mathrm{2}} \\ $$$$=\mathrm{2sin}\:\frac{{A}+{B}}{\mathrm{2}}\left\{\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}−\mathrm{cos}\:\frac{{A}+{B}+\mathrm{2}{C}}{\mathrm{2}}\right\} \\ $$$$=\mathrm{2sin}\:\frac{{A}+{B}}{\mathrm{2}}.\mathrm{2sin}\:\:\frac{{A}+{C}}{\mathrm{2}}\mathrm{sin}\:\frac{{B}+{C}}{\mathrm{2}} \\ $$

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