`a)` Sửa đề `\sqrt{2}sin(x+π/4)`
Ta có:
`\qquad \sqrt{2}sin(x+π/4)`
`=\sqrt{2}. (sinxcos\ π/ 4 +cosx sin\ π/4)`
`=\sqrt{2}.(sinx. 1/{\sqrt{2}}+cosx. 1/{\sqrt{2}})`
`=sinx+cosx` $\quad (1)$
$\\$
`\qquad \sqrt{2}cos(x-π/4)`
`=\sqrt{2}. (cosxcos\ π/ 4 +sinx sin\ π/4`
`=\sqrt{2}.(cosx. 1/{\sqrt{2}}+sinx. 1/{\sqrt{2}})`
`=cosx+sinx`
`=sinx+cosx` $\quad (2)$
$\\$
Từ `(1);(2)` suy ra:
`\qquad sinx+cosx`
`=\sqrt{2}. sin(x+π/4)`
`=\sqrt{2}. cos(x-π/4)`
$\\$
`b)` `sinx+ √3.cosx=2sin(x+π/ 3)`
Ta có:
`VP= 2sin(x+π/3)`
`=2 . (sinxcos \ π/3+cosx.sin\ π/3)`
`=2.(sinx. 1/ 2 +cosx . \sqrt{3}/2)`
`=sinx+\sqrt{3}cosx=VT`
Vậy `sinx+\sqrt{3}cosx=2sin(x+π/3)`
