Question Number 209 by 02@>@0 last updated on 25/Jan/15
![y=sin(x)+cos(x)](https://www.tinkutara.com/question/Q209.png)
$${y}={sin}\left({x}\right)+{cos}\left({x}\right) \\ $$
Answered by 123456 last updated on 15/Dec/14
![y=((√2)/( (√2)))sin x+((√2)/( (√2)))cos x y=(√2)((1/( (√2)))sin x+(1/( (√2)))cos x) y=(√2)(cos (π/4)sin x+sin (π/4)cos x) y=(√2)sin (x+(π/4))](https://www.tinkutara.com/question/Q212.png)
$${y}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}\mathrm{sin}\:{x}+\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}\mathrm{cos}\:{x} \\ $$$${y}=\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{cos}\:{x}\right) \\ $$$${y}=\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\mathrm{sin}\:{x}+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\mathrm{cos}\:{x}\right) \\ $$$${y}=\sqrt{\mathrm{2}}\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right) \\ $$