f(x)=cos(logx)show that f(1/x)f(1/y)-1/2×{f(x/y)+f(xy)}=0
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f(x) = cos(logx)
f(1/x) = cos(log1/x) = cos(logx^-1) =cos(-logx)
= cos(logx) [ because cos(-∅) = cos∅]
similarly , f(y) = cos(logy)
f(x/y) = cos(logx/y) = cos(logx - logy)
f(xy) = cos(logxy) = cos(logx + logy)
now, f(x/y) + f(xy) = 2cos(logx).cos(logy)
[ because cos(A - B) = cosA.cosB+sinA.sinB and cos(A + B) = cosA.cosB-sinA.sinB ]
now, f(1/x)(1/y) - 1/2[f(x/y) + f(xy)]
= cos(logx).cos(logy) - 1/2[2cos(logx).cos(logy)]
= cos(logx).cos(logy) - cos(logx).cos(logy)
= 0 hence, proved
f(1/x) = cos(log1/x) = cos(logx^-1) =cos(-logx)
= cos(logx) [ because cos(-∅) = cos∅]
similarly , f(y) = cos(logy)
f(x/y) = cos(logx/y) = cos(logx - logy)
f(xy) = cos(logxy) = cos(logx + logy)
now, f(x/y) + f(xy) = 2cos(logx).cos(logy)
[ because cos(A - B) = cosA.cosB+sinA.sinB and cos(A + B) = cosA.cosB-sinA.sinB ]
now, f(1/x)(1/y) - 1/2[f(x/y) + f(xy)]
= cos(logx).cos(logy) - 1/2[2cos(logx).cos(logy)]
= cos(logx).cos(logy) - cos(logx).cos(logy)
= 0 hence, proved
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