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Identity? 'Show that sin(2^5x) = 2^5sinxcosx...'
- Thread starter messa
- Start date
Re: Identity?
Hello, messa!
The "..." means what it always means: "and so on, in the same manner."
(In this problem, it's really stupid ... since they omitted just one term.) **
It says: \(\displaystyle \:\sin(32x) \:=\:32\cdot\sin x\cdot\cos x\cdot\cos2x\cdot\cos4x\cdot\cos8x\cdot\cos16x\)
We use the half-angle identity: \(\displaystyle \:\sin(2\theta)\:=\:2\cdot\sin\theta\cdot\cos\theta\) . . . over and over.
We have: \(\displaystyle \;\sin(32x) \:=\:2\cdot\underbrace{\sin16x}\cdot\cos16x\)
. . . . . . . . . . . \(\displaystyle =\:2\cdot\overbrace{(2\cdot\sin8x\cdot \cos8x)}\cdot\cos16x\)
. . . . . . . . . . . \(\displaystyle =\:4\cdot\underbrace{\sin8x}\cdot \cos8x\cdot\cos16x\)
. . . . . . . .\(\displaystyle = \:4\overbrace{(2\cdot\sin4x\cdot\cos4x)}\cdot\cos8x\cdot\cos16x\)
. . . . . . . .\(\displaystyle =\:8\cdot\underbrace{\sin4x}\cdot\cos4x \cdot\cos8x \cdot\cos16x\)
. . . . .\(\displaystyle = \;8\overbrace{(2\cdot\sin2x\cdot\cos2x)}\cdot\cos4x\cdot\cos8x\cdot16x\)
. . . . .\(\displaystyle = \:16\cdot\underbrace{\sin2x}\cdot\cos2x\cdot \cos4x\cdot \cos8x \cdot\cos16x\)
. . \(\displaystyle = \:16\overbrace{(2\cdot\sin x\cdot\cos x)}\cdot \cos2x\cdot\cos4x\cdot\cos8x\cdot\cos16x\)
. . \(\displaystyle =\:32\cdot\sin x\cdot\cos x\cdot\cos2x\cdot\cos4x\cdot\cos8x\cdot\cos16x\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
Suppose I wrote: \(\displaystyle \:2,\,4,\,6,\,8,\,10,\,12,\,14,\,16,\,\cdots\) and so on up to \(\displaystyle 20.\)
It would be simpler (and smarter) to simply write the "18".
. . Pretty stupid, right?
Hello, messa!
Show that: \(\displaystyle \:\sin(2^5x) \:= \:2^5\sin x\cos x\cos2x\cos4x\,\cdots\,\cos16x\)
Would I use the half angle identity to start?
And what does the ... mean?
The "..." means what it always means: "and so on, in the same manner."
(In this problem, it's really stupid ... since they omitted just one term.) **
It says: \(\displaystyle \:\sin(32x) \:=\:32\cdot\sin x\cdot\cos x\cdot\cos2x\cdot\cos4x\cdot\cos8x\cdot\cos16x\)
We use the half-angle identity: \(\displaystyle \:\sin(2\theta)\:=\:2\cdot\sin\theta\cdot\cos\theta\) . . . over and over.
We have: \(\displaystyle \;\sin(32x) \:=\:2\cdot\underbrace{\sin16x}\cdot\cos16x\)
. . . . . . . . . . . \(\displaystyle =\:2\cdot\overbrace{(2\cdot\sin8x\cdot \cos8x)}\cdot\cos16x\)
. . . . . . . . . . . \(\displaystyle =\:4\cdot\underbrace{\sin8x}\cdot \cos8x\cdot\cos16x\)
. . . . . . . .\(\displaystyle = \:4\overbrace{(2\cdot\sin4x\cdot\cos4x)}\cdot\cos8x\cdot\cos16x\)
. . . . . . . .\(\displaystyle =\:8\cdot\underbrace{\sin4x}\cdot\cos4x \cdot\cos8x \cdot\cos16x\)
. . . . .\(\displaystyle = \;8\overbrace{(2\cdot\sin2x\cdot\cos2x)}\cdot\cos4x\cdot\cos8x\cdot16x\)
. . . . .\(\displaystyle = \:16\cdot\underbrace{\sin2x}\cdot\cos2x\cdot \cos4x\cdot \cos8x \cdot\cos16x\)
. . \(\displaystyle = \:16\overbrace{(2\cdot\sin x\cdot\cos x)}\cdot \cos2x\cdot\cos4x\cdot\cos8x\cdot\cos16x\)
. . \(\displaystyle =\:32\cdot\sin x\cdot\cos x\cdot\cos2x\cdot\cos4x\cdot\cos8x\cdot\cos16x\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
Suppose I wrote: \(\displaystyle \:2,\,4,\,6,\,8,\,10,\,12,\,14,\,16,\,\cdots\) and so on up to \(\displaystyle 20.\)
It would be simpler (and smarter) to simply write the "18".
. . Pretty stupid, right?