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Prove that:`(2cos2^ntheta+1)/(2costheta+1)=(2costheta-1)(2cos2theta-1)(2cos2^2theta-1)(2cos2^(n-1)theta-1)` |
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Answer» `R.H.S. = (2costheta-1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)` `=1/(2costheta+1)[(2costheta+1)(2costheta-1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)]` `=1/(2costheta+1)[(4cos^2theta-1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)]` `=1/(2costheta+1)[(2(2cos^2theta)-1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)]` `=1/(2costheta+1)[(2(1+cos2theta)-1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)]` `=1/(2costheta+1)[(2cos2theta+1)(2cos2theta-1)...(2cos2^(n-1)theta - 1)]` `=1/(2costheta+1)[(4cos^ 2theta -1)...(2cos2^(n-1)theta - 1)]` `=1/(2costheta+1)[(2cos4theta+1)(2cos4theta-1)...(2cos2^(n-1)theta - 1)]` Similarly, if we solve this, it will become, `=1/(2costheta+1)[(2cos2^(n-1)theta + 1)(2cos2^(n-1)theta - 1)]` `=1/(2costheta+1)[4cos^2 2^(n-1)theta-1]` `=1/(2costheta+1)[2(cos2^n theta+1)-1]` `=(2cos2^n theta+1)/(2costheta+1) = R.H.S.` |
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