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सिद्ध कीजिए कि `cos 6x=32cos^(6)x-48cos^(4)x+18cos^(2)x-1` |
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Answer» बायाँ पक्ष `=cos6x` `=cos 3(2x)=cos3theta=cos(2theta+theta)` `=cos 2 theta cos theta - sin 2 theta sin theta" "("माना "theta = 2x)` `=(2cos^(2)theta-1)cos theta - 2 sin theta cos theta sin theta` `=2cos^(3)theta-cos theta-2cos theta(1-cos^(2)theta)` `=2cos^(3)theta-cos theta-2cos theta+2cos^(3)theta` `=4cos^(3)theta-3cos theta =4cos^(3)2x-3 cos 2x" "(because theta=2x)` `=4(2cos^(2)x-1)^(3)-3(2cos^(2)x-1)` `=4(8cos^(6)x-12 cos^(4)x+6cos^(2)x-1)-(6cos^(2)x-3)` `=32cos^(6)x-48cos^(4)x+18cos^(2)x-1=` दायाँ पक्ष |
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