# Help Needed with this Trigonometric Proof

• Feb 3rd 2011, 09:58 PM
sammysparrow
Help Needed with this Trigonometric Proof
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!
• Feb 3rd 2011, 10:06 PM
harish21
$\displaystyle RHS = \sqrt{2}\;\bigg[ \sin\bigg(A+\dfrac{\pi}{4}\bigg)\bigg]$

now use $\displaystyle \sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$ where $\displaystyle B = \dfrac{\pi}{4}$
• Feb 4th 2011, 07:13 AM
Soroban
Hello, sammysparrow!

Quote:

$\displaystyle \text{Prove that: }\:\sin A + \cos A \:=\:\sqrt{2}\sin\left(A + \frac{\pi}{4}\right)$

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: .$\displaystyle \sqrt{2}\left(\sin A\cos\frac{\pi}{4} + \cos A\sin\frac{\pi}{4}\right) \;=\;\sqrt{2}\sin(A +\frac{\pi}{4})$

• Feb 4th 2011, 12:55 PM
Quote:

Originally Posted by sammysparrow
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!

If you had to begin from the LHS, another way is

$\displaystyle 2SinACosB=Sin(A+B)+Sin(A-B)$

$\displaystyle 2CosASinB=Sin(A+B)-Sin(A-B)$

$\displaystyle \Rightarrow\ 2(SinACosB+CosASinB)=2Sin(A+B)$

$\displaystyle \Rightarrow\ SinACosB+CosASinB=Sin(A+B)$

$\displaystyle SinACosB+CosASinB=k(SinA+CosA)$

$\displaystyle \Rightarrow\ SinB=CosB\Rightarrow\ B=\frac{\pi}{4}$

$\displaystyle \displaystyle\ Sin\frac{\pi}{4}=Cos\frac{\pi}{4}=\frac{1}{\sqrt{2 }}$

$\displaystyle \Rightarrow\ SinA+CosA=\sqrt{2}\left[SinA+\frac{\pi}{4}\right]$
• Feb 6th 2011, 06:05 PM
sammysparrow
(Hi) Thanks everyone!