1.y=(sinx/4)^4+(cosx/4)^4y'=4(sinx/4)^3{(1/4)cosx/4)+4(cosx/4)^3{(-1/4)sinx/4)=(sinx/4)^3(cosx/4)-sinx/4(cosx/4)^3=sinx/4cosx/4[(sinx/4)^2-(cosx/4)^2]=(-1/2)sinx/2cosx/2=(-1/4)sinx从后面数来第二条式子是怎样从它前

Question

1.y = (sinx/4)^4 + (cosx/4)^4
y' = 4(sinx/4)^3 {(1/4)cosx/4) + 4(cosx/4)^3 {(-1/4)sinx/4)
= (sinx/4)^3(cosx/4) - sinx/4(cosx/4)^3
= sinx/4cosx/4[ (sinx/4)^2 - (cosx/4)^2]
= (-1/2) sinx/2cosx/2
= (-1/4) sinx
从后面数来第二条式子是怎样从它前面那条式子得来的?

Answer 1

(sinx/4)^2 - (cosx/4)^2=-cosx/2
sinx/4cosx/4=1/2sinx/2
两个都是二倍角公式的逆用