[cardinopolynomial]

4.ได้$ y= (sinx+cosx)(1-sinxcosx)$

$= (\sqrt{2})(\frac{1}{2}) $

$z=cos(2tan^{-1}(\frac{1}{\sqrt{2}}))$

ลองมอง $A= 2tan^{-1}(\frac{1}{\sqrt{2}})$

จากเอกลักษณ์ $(secA)^2-(tanA)^2=1$

$(secA)^2=1+(tanA)^2$

$tanA=tan(2tan^{-1}(\frac{1}{\sqrt{2}}))$

$=tan2B=\frac{2tanB}{1-(tanB)^2};tan^{-1}(\frac{1}{\sqrt{2}})=B $

$=\frac{2x\frac{1}{\sqrt{2}}}{1-\frac{1}{2}}=2\sqrt{2} $

$secA=3$

$cos(2tan^{-1}(\frac{1}{\sqrt{2}}))=\frac{1}{3} $