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Trig Proof. Killing me!
- Thread starter fran33s
- Start date
Hello, fran33s!
We need two sum-to-product identities:
. . \(\displaystyle \begin{array}{ccc}\sin A + \sin B &=& 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \\ \cos A + \cos B &=& 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\end{array}\)
The left side is: .\(\displaystyle \underbrace{(\sin7x + \sin x)} + \underbrace{(\sin5x+\sin3x)}\)
. . . . . . . . . . \(\displaystyle =\;\;\overbrace{2\sin4x\cos3x} \;+\; \overbrace{2\sin4x\cos x}\)
. . . . . . . . . . \(\displaystyle =\;\;2\sin4x\underbrace{(\cos3x + \cos x)}\)
. . . . . . . . . . \(\displaystyle =\;\;2\sin4x\cdot \overbrace{2\cos2x\cos x}\)
. . . . . . . . . . \(\displaystyle =\;\;2\cos x\cos2x\sin4x\)
We need two sum-to-product identities:
. . \(\displaystyle \begin{array}{ccc}\sin A + \sin B &=& 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \\ \cos A + \cos B &=& 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\end{array}\)
\(\displaystyle \sin x + \sin 3x + \sin 5x + \sin 7x \:=\: 4\cos x\cos2x\sin4x\)
The left side is: .\(\displaystyle \underbrace{(\sin7x + \sin x)} + \underbrace{(\sin5x+\sin3x)}\)
. . . . . . . . . . \(\displaystyle =\;\;\overbrace{2\sin4x\cos3x} \;+\; \overbrace{2\sin4x\cos x}\)
. . . . . . . . . . \(\displaystyle =\;\;2\sin4x\underbrace{(\cos3x + \cos x)}\)
. . . . . . . . . . \(\displaystyle =\;\;2\sin4x\cdot \overbrace{2\cos2x\cos x}\)
. . . . . . . . . . \(\displaystyle =\;\;2\cos x\cos2x\sin4x\)
Hello, allh5!
Don't tack your problem on the end of someone else's post.
You're lucky that I read this post again . . .
By the way, that Greek letter is pi.
Double-angle identity: .\(\displaystyle \sin2\theta \:=\:2\sin\theta\cos\theta\)
So we have:[COLOR=#e00e] .[/COLOR]\(\displaystyle \sin3x \:=\:2\cos2x\)
... \(\displaystyle 2\sin2x\cos2x \:=\:2\cos2x\)
. . \(\displaystyle 2\sin2x\cos2x - 2\cos2x \:=\:0\)
. . \(\displaystyle 2\cos2x(\sin2x - 1) \:=\:0\)
Then: .\(\displaystyle \cos2x \:=\:0 \quad\Rightarrow\quad 2x \:=\:\frac{\pi}{2} + \pi n \quad\Rightarrow\quad x \:=\:\frac{\pi}{4} + \frac{\pi}{2}n \)
. . \(\displaystyle \sin2x-1\:=\:0 \quad\Rightarrow\quad \sin2x \:=\:1 \quad\Rightarrow\quad 2x \:=\:\frac{\pi}{2} + 2\pi n\) . \(\displaystyle \Rightarrow\quad x \:=\:\frac{\pi}{4} + \pi n \)
Therefore: .\(\displaystyle \boxed{x \;=\;\frac{\pi}{4},\:\frac{3\pi}{4},\:\frac{5\pi}{4},\:\frac{7\pi}{4}}\)
Don't tack your problem on the end of someone else's post.
You're lucky that I read this post again . . .
By the way, that Greek letter is pi.
\(\displaystyle \text{Solve: }\:\sin 4x\:=\:2\cos2x,\;\;[0,\,2\pi)\)
Double-angle identity: .\(\displaystyle \sin2\theta \:=\:2\sin\theta\cos\theta\)
So we have:[COLOR=#e00e] .[/COLOR]\(\displaystyle \sin3x \:=\:2\cos2x\)
... \(\displaystyle 2\sin2x\cos2x \:=\:2\cos2x\)
. . \(\displaystyle 2\sin2x\cos2x - 2\cos2x \:=\:0\)
. . \(\displaystyle 2\cos2x(\sin2x - 1) \:=\:0\)
Then: .\(\displaystyle \cos2x \:=\:0 \quad\Rightarrow\quad 2x \:=\:\frac{\pi}{2} + \pi n \quad\Rightarrow\quad x \:=\:\frac{\pi}{4} + \frac{\pi}{2}n \)
. . \(\displaystyle \sin2x-1\:=\:0 \quad\Rightarrow\quad \sin2x \:=\:1 \quad\Rightarrow\quad 2x \:=\:\frac{\pi}{2} + 2\pi n\) . \(\displaystyle \Rightarrow\quad x \:=\:\frac{\pi}{4} + \pi n \)
Therefore: .\(\displaystyle \boxed{x \;=\;\frac{\pi}{4},\:\frac{3\pi}{4},\:\frac{5\pi}{4},\:\frac{7\pi}{4}}\)