If cosx-sinx= root2 sinx, prove that cosx+sinx= 2 cosx.
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cosx-sinx=√2sinx
or, cosx=√2sinx+sinx
or, cosx=sinx(√2+1)
or, cosx=sinx(√2+1)(√2-1)/(√2-1)
or, (√2-1)cosx=sinx{(√2)²-(1)²}
or, √2cosx-cosx=sinx(2-1)
or, √2cosx-cosx=sinx
or, -cosx-sinx=-√2cosx
or, cosx+sinx=√2cosx
or, cosx=√2sinx+sinx
or, cosx=sinx(√2+1)
or, cosx=sinx(√2+1)(√2-1)/(√2-1)
or, (√2-1)cosx=sinx{(√2)²-(1)²}
or, √2cosx-cosx=sinx(2-1)
or, √2cosx-cosx=sinx
or, -cosx-sinx=-√2cosx
or, cosx+sinx=√2cosx
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