## Trig Induction: show sin(2^n @) = 2^n sin(@)cos(@)cos(2@)...

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

### Trig Induction: show sin(2^n @) = 2^n sin(@)cos(@)cos(2@)...

Using an inductive argument to show that $sin(2^n\theta) = 2^nsin\theta cos\theta cos2\theta...cos(2^{n-1}\theta)$

My attempts so far are:
Step 1. See if it's true for n=1. So,
$sin(2^1\theta) = 2^1sin\theta cos\theta cos2\theta... cos(2^{1-1}\theta)$
$sin(2\theta) = 2sin\theta cos\theta cos2\theta...cos\theta$
So LHS = RHS (I think I can say this based on the identity sin2A=2sinAcosA? This is my first issue)

Step 2. Assume it holds true for n = k
So, $sin(2^k\theta) = 2^ksin\theta cos\theta cos2\theta... cos(2^{k-1}\theta)$

Step 3. Show it's true for k+1
Here I know if it is true, then
$sin(2^{k+1}\theta) = 2^{k+1}sin\theta cos\theta cos2\theta... cos(2^{k+1-1}\theta)$
$=2^{k+1}sin\theta cos\theta cos2\theta... cos(2^k\theta)$

So taking the LHS $sin(2^{k+1}\theta$
$=sin(2^k.2^1.\theta$ Using index laws
$=2sin(2^k\theta)cos(2^k\theta)$ Using sin2A = 2sinAcosA

Now here is where I am confused. I know back in Step 2 I have $sin(2^k\theta) = 2^ksin\theta cos\theta cos2\theta... cos(2^{k-1}\theta)$ and need to substitute this in but I'm unsure how to resolve it to get the answer I know I need!

Hope this all makes sense and that someone can give me a hand. Thanks very much.
Nar

Posts: 6
Joined: Sat Jan 04, 2014 2:34 am

### Re: Trig Induction: show sin(2^n @) = 2^n sin(@)cos(@)cos(2@

Nar wrote:Using an inductive argument to show that $sin(2^n\theta) = 2^nsin\theta cos\theta cos2\theta...cos(2^{n-1}\theta)$

Step 1. See if it's true for n=1.

For n = 1, 2^n = 2^1 = 2, 2^(n-1) = 2^0 = 1, and:

LHS: sin(2^n @) = sin(2^1 @) = sin(2@) = 2sin(@)cos(@)
RHS: 2^n sin(@)cos(@)...cos(2^(n-1)@) = 2^1 sin(@)cos(@)...cos(2^0 @) = 2sin(@)cos(@)

(ecause cos(2^(n-2)@) = cos(@))

Nar wrote:Step 2. Assume it holds true for n = k
So, $sin(2^k\theta) = 2^ksin\theta cos\theta cos2\theta... cos(2^{k-1}\theta)$

Step 3. Show it's true for k+1
Here I know if it is true, then
$sin(2^{k+1}\theta) = 2^{k+1}sin\theta cos\theta cos2\theta... cos(2^{k+1-1}\theta)$
$=2^{k+1}sin\theta cos\theta cos2\theta... cos(2^k\theta)$

So taking the LHS $sin(2^{k+1}\theta$
$=sin(2^k.2^1.\theta$ Using index laws
$=2sin(2^k\theta)cos(2^k\theta)$ Using sin2A = 2sinAcosA

Now here is where I am confused. I know back in Step 2 I have $sin(2^k\theta) = 2^ksin\theta cos\theta cos2\theta... cos(2^{k-1}\theta)$ and need to substitute this in but I'm unsure how to resolve it....

Plug it in:

LHS: sin(2^(k+1) @ = sin(2 (2^k @)) = 2sin(2^k @)cos(2^k @)

substitute:

LHS: = 2[2^k sin(@)cos(@)cos(2@)...cos(2^(k-1) @)] cos(2^k @) = (2)(2^k)sin(@)cos(@)cos(2@)...cos(2^(k-1) @) cos(2^k @)

Now do the last step.

maggiemagnet

Posts: 298
Joined: Mon Dec 08, 2008 12:32 am

### Re: Trig Induction: show sin(2^n @) = 2^n sin(@)cos(@)cos(2@

OK thanks very much, just 2 questions...

(ecause cos(2^(n-2)@) = cos(@))

Is this an identity, I don't know?

FInal step
$(2)(2^ksin\theta cos\theta cos2\theta...cos(2^{k-1}\theta)) cos(2^k\theta$

I know I can use index law for the first part
$2^1.2^ksin\theta...... = 2^{k+1}\theta$

But I don't know what to do with the .....$cos(2^{k-1}\theta).cos(2^k\theta)$ at the end

Thanks very much, I think I'm missing a very basic concept here!
Nar

Posts: 6
Joined: Sat Jan 04, 2014 2:34 am

### Re: Trig Induction: show sin(2^n @) = 2^n sin(@)cos(@)cos(2@

Nar wrote:
(because cos(2^(n-2)@) = cos(@))

Is this an identity, I don't know?

It's plugging "1" in for "n" in the formula and then simplifying.

Nar wrote:But I don't know what to do with the .....$cos(2^{k-1}\theta).cos(2^k\theta)$ at the end

I plugged into the formula and then simplified.

maggiemagnet

Posts: 298
Joined: Mon Dec 08, 2008 12:32 am

### Re: Trig Induction: show sin(2^n @) = 2^n sin(@)cos(@)cos(2@

I'm fine with this all until the very last step. I can't see how the following simplifies

$(2)(2^ksin\theta cos\theta cos2\theta...cos(2^{k-1}\theta)) cos(2^k\theta) = 2^{k+1}sin\theta cos\theta 2cos\theta... cos(2^k\theta)$

I know the $2.2^k$ at the front becomes $2^{k+1}$ but what happens with the both cos terms at the end?
Nar

Posts: 6
Joined: Sat Jan 04, 2014 2:34 am

Nar wrote:I'm fine with this all until the very last step. I can't see how the following simplifies

$(2)(2^ksin\theta cos\theta cos2\theta...cos(2^{k-1}\theta)) cos(2^k\theta) = 2^{k+1}sin\theta cos\theta 2cos\theta... cos(2^k\theta)$

...what happens with the both cos terms at the end?

Nothing "happens" with the cosines. The brackets are removed. What did you get when you wrote everything out and removed the brackets?

stapel_eliz

Posts: 1708
Joined: Mon Dec 08, 2008 4:22 pm