Hi, can someone show me how to prove that
sin A + cos A = (the square root of 2) sin ( A + pi over 4) ?
Thanks, any help would be very much appreciated!
Hello, sammysparrow!
$\displaystyle \text{Prove that: }\:\sin A + \cos A \:=\:\sqrt{2}\sin\left(A + \frac{\pi}{4}\right)$
Multiply the left side by $\displaystyle \frac{\sqrt{2}}{\sqrt{2}}$
. . $\displaystyle \frac{\sqrt{2}}{\sqrt{2}}\left(\sin A + \cos A\right) \;=\;\sqrt{2}\left(\sin A\cdot\frac{1}{\sqrt{2}}+ \cos A\cdot\frac{1}{\sqrt{2}}\right) $ .[1]
Note that: .$\displaystyle \sin\frac{\pi}{4} \,=\,\cos\frac{\pi}{4} \,=\,\frac{1}{\sqrt{2}}$
Then [1] becomes: .$\displaystyle \sqrt{2}\left(\sin A\cos\frac{\pi}{4} + \cos A\sin\frac{\pi}{4}\right) \;=\;\sqrt{2}\sin(A +\frac{\pi}{4}) $
If you had to begin from the LHS, another way is
$\displaystyle 2SinACosB=Sin(A+B)+Sin(A-B)$
$\displaystyle 2CosASinB=Sin(A+B)-Sin(A-B)$
$\displaystyle \Rightarrow\ 2(SinACosB+CosASinB)=2Sin(A+B)$
$\displaystyle \Rightarrow\ SinACosB+CosASinB=Sin(A+B)$
$\displaystyle SinACosB+CosASinB=k(SinA+CosA)$
$\displaystyle \Rightarrow\ SinB=CosB\Rightarrow\ B=\frac{\pi}{4}$
$\displaystyle \displaystyle\ Sin\frac{\pi}{4}=Cos\frac{\pi}{4}=\frac{1}{\sqrt{2 }}$
$\displaystyle \Rightarrow\ SinA+CosA=\sqrt{2}\left[SinA+\frac{\pi}{4}\right]$