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The expression `[(1+sin(pi/8)+icos(pi/8))/(1+sin(pi/8)-icos(pi/8))]^8` is equal isA. 1B. `-1`C. iD. `-i` |
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Answer» Correct Answer - B Let `sin""(pi)/(8)+icos""(pi)/(8)=z` `rArr [(1+sin""(pi)/(8)+icos""(pi)/(8))/(1+sin""(pi)/(8)-icos""(pi)/(8))]^(8)` `=((1+z)/(1+(1)/(z)))^(8)=z^(8)` `=(sin""(pi)/(8)+icos""(pi)/(8))^(8)` `=(cos""((pi)/(2)-(pi)/8)+isin""((pi)/(2)-(pi)/(8)))^(8)` `=(cos""(3pi)/(8)+isin""(3pi)/(8))^(8)=cos3pi=-1` |
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