A) \[1-4cos\text{ }A\text{ }cos\text{ }Bcos\text{ }C\]
B) \[1+2\text{ }cos\text{ }A\text{ }cos\text{ }Bcos\text{ }C\]
C) \[1-4\sin Asin\,BsinC\]
D) \[4\text{ }sin\text{ }A\text{ }sin\text{ }B\text{ }sin\text{ }C\]
Correct Answer: C
Solution :
Given, \[A+B+C=\frac{3\pi }{2}\] \[\therefore \] \[cos\text{ }2A+cos\text{ }2B+cos\text{ }2C\] \[=2cos\left( \frac{2A+2B}{2} \right)\cos \left( \frac{2A-2B}{2} \right)+\cos 2C\] \[=2\cos (A+B)\cos (A-B)+\cos 2C\] \[=2\cos \left( \frac{3\pi }{2}-C \right)\cos (A-B)+1-2{{\sin }^{2}}C\] \[=-2\sin C\cos (A-B)+1-2{{\sin }^{2}}C\] \[\sin \left( \frac{3\pi }{2}-(A+B) \right)\] \[=1-2\sin C\left[ \cos (A-B)+\sin \left( \frac{3\pi }{2}-(A+B) \right) \right]\] \[=1-2\sin C[\cos (A-B)-cos(A+B)]\] \[=1-2\sin C[2\sin A\sin B]\] \[=1-4\sin A\sin B\sin C\]
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