# How do you derive the following trig identity?

• Jan 30th 2009, 02:13 PM
Kaitosan
How do you derive the following trig identity?
• Jan 30th 2009, 03:34 PM
Soroban
Hello, Kaitosan!

Quote:

$\sin x + \cos x \:=\:\sqrt{2}\,\sin(x + \tfrac{\pi}{4})$

$\sqrt{2}\,\sin(x + \tfrac{\pi}{4}) \;=\;\sqrt{2}\left(\sin x\cos\tfrac{\pi}{4} + \cos x\sin\tfrac{\pi}{4}\right)$

. . . . . . . . . $= \;\sqrt{2}\,\bigg[\sin x\cdot\frac{1}{\sqrt{2}} + \cos x\cdot\frac{1}{\sqrt{2}}\bigg]$

. . . . . . . . . $= \;\sin x + \cos x$

• Jan 30th 2009, 09:39 PM
Kaitosan
Actually... I'd rather you try with the left side (I have specific reasons). Can you do that....?

This is as far as I can go -

(sinx)^2 + (cosx)^2 = 1

(sinx + cosx)^2 = 1 - sin2x

sinx + cosx = sqrt(1 - sin2x)

.................................................. ............
• Jan 31st 2009, 01:39 AM
running-gag
Hi

This is not the good way to proceed.

$a \sin x + b \cos x = \sqrt{a^2+b^2} \left(\frac{a}{\sqrt{a^2+b^2}} \sin x + \frac{b}{\sqrt{a^2+b^2}} \cos x \right)$

$\frac{a}{\sqrt{a^2+b^2}}$ and $\frac{b}{\sqrt{a^2+b^2}}$ can be seen as the cos and the sin of an angle $\theta$ because the sum of their square is equal to 1

$a \sin x + b \cos x = \sqrt{a^2+b^2} \left(\sin x \cos \theta+ \cos x \sin \theta\right) = \sqrt{a^2+b^2} \:\sin (x + \theta)$
• Jan 31st 2009, 02:02 AM
red_dog
$\sin x+\cos x=\sin x+\sin\left(\frac{\pi}{2}-x\right)=$

$=2\sin\frac{x+\frac{\pi}{2}-x}{2}\cos\frac{x-\frac{\pi}{2}+x}{2}=$

$=2\sin\frac{\pi}{4}\cos\left(x-\frac{\pi}{4}\right)=$

$=2\frac{\sqrt{2}}{2}\cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}\sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=$

$=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$
• Jan 31st 2009, 02:10 AM
Rincewind
Quote:

Originally Posted by Kaitosan
Actually... I'd rather you try with the left side (I have specific reasons). Can you do that....?

This is as far as I can go -

(sinx)^2 + (cosx)^2 = 1

(sinx + cosx)^2 = 1 - sin2x

sinx + cosx = sqrt(1 - sin2x)

.................................................. ............

Can't you just take Soroban's answer and play it in reverse.

For example. First take $\sqrt{2}$ out as a factor

$\sin x + \cos x = \sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x\right)$

Then observe that $\sin \frac{\pi}{4} = \cos \frac{\pi}{4}= \frac{1}{\sqrt{2}}$, and therefore

$\qquad = \sqrt{2}\left(\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}\right)$

And then from the compound angle formula for sin

$\qquad = \sqrt{2}\sin \left(x + \frac{\pi}{4}\right)$
• Jan 31st 2009, 12:07 PM
Kaitosan
Quote:

Originally Posted by red_dog
$\sin x+\cos x=\sin x+\sin\left(\frac{\pi}{2}-x\right)=$

$=2\sin\frac{x+\frac{\pi}{2}-x}{2}\cos\frac{x-\frac{\pi}{2}+x}{2}=$

$=2\sin\frac{\pi}{4}\cos\left(x-\frac{\pi}{4}\right)=$

$=2\frac{\sqrt{2}}{2}\cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}\sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=$

$=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$

Wow. Um... I'm completely lost on the second step. Hmm.
• Jan 31st 2009, 12:48 PM
red_dog
In the second step I used the identity

$\sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$
• Jan 31st 2009, 01:22 PM
Kaitosan
Interesting! Could you please explain how that identity comes into being because I've never seen it? Right now I'm in calculus and I just want to make sure I know all the identities.