Let s(x)=(2^x-2^-x)/2 and c(x)=(2^x+2^-x)/2. Compute

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Pembahasan

Untuk menghitung [c(x)]^2 - [s(x)]^2, kita substitusikan definisi c(x) dan s(x):

c(x) = (2^x + 2^-x) / 2
s(x) = (2^x - 2^-x) / 2

[c(x)]^2 = [ (2^x + 2^-x) / 2 ]^2 = ( (2^x)^2 + 2(2^x)(2^-x) + (2^-x)^2 ) / 4 = (2^{2x} + 2(1) + 2^{-2x}) / 4 = (2^{2x} + 2 + 2^{-2x}) / 4

[s(x)]^2 = [ (2^x - 2^-x) / 2 ]^2 = ( (2^x)^2 - 2(2^x)(2^-x) + (2^-x)^2 ) / 4 = (2^{2x} - 2(1) + 2^{-2x}) / 4 = (2^{2x} - 2 + 2^{-2x}) / 4

[c(x)]^2 - [s(x)]^2 = [ (2^{2x} + 2 + 2^{-2x}) / 4 ] - [ (2^{2x} - 2 + 2^{-2x}) / 4 ]
= ( (2^{2x} + 2 + 2^{-2x}) - (2^{2x} - 2 + 2^{-2x}) ) / 4
= ( 2^{2x} + 2 + 2^{-2x} - 2^{2x} + 2 - 2^{-2x} ) / 4
= (2 + 2) / 4
= 4 / 4
= 1

Jadi, [c(x)]^2 - [s(x)]^2 = 1.