A) \[-\cos \,3x\]
B) \[-\sin \,3x\]
C) \[\sin \,3x\]
D) \[\cos \,3x\]
Correct Answer: B
Solution :
\[{{\left( sin\,x+i\,cos\,x \right)}^{3}}\] \[= si{{n}^{3}}x + {{(i)}^{3}} co{{s}^{3}} x + 3i \left( sin x \right) \left( cos x \right)\] \[\left( sin x + i cos x \right)\] \[= sin x- i co{{s}^{3}} x + 3i si{{n}^{2}} x cos x- 3 sin x co{{s}^{2}} x\] \[=si{{n}^{3}} x-3sin\,x\,co{{s}^{2}} x+i\,cos\,x\left( co{{s}^{2}} x+si{{n}^{2}} x \right)\] \[=sin\,x\left( si{{n}^{2}} x-3\,co{{s}^{2}}x \right)+i\,cos\,x\] Real part of \[{{\left( sin x + i cos x \right)}^{3}}\] \[= sin x \left( si{{n}^{2}} x - 3 co{{s}^{2}} x \right)\] \[=\,\,\,sin\,x\left[ si{{n}^{2}} x-3\left( 1-si{{n}^{2}}\,x \right) \right]\] \[= sin x \left[ 4 si{{n}^{2}}\,x - 3 \right]\] \[=\,\,\,4 si{{n}^{3}} x-3\,\,sin\,x\] \[= -\left( 3 sin x - 4 si{{n}^{3}} x \right)\] \[=-sin3x\]
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