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

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

Postby Nar on Sun Jan 05, 2014 1:34 am

Using an inductive argument to show that

My attempts so far are:
Step 1. See if it's true for n=1. So,


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,

Step 3. Show it's true for k+1
Here I know if it is true, then



So taking the LHS
Using index laws
Using sin2A = 2sinAcosA

Now here is where I am confused. I know back in Step 2 I have 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.
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Re: Trig Induction: show sin(2^n @) = 2^n sin(@)cos(@)cos(2@

Postby maggiemagnet on Sun Jan 05, 2014 6:39 pm

Nar wrote:Using an inductive argument to show that

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,

Step 3. Show it's true for k+1
Here I know if it is true, then



So taking the LHS
Using index laws
Using sin2A = 2sinAcosA

Now here is where I am confused. I know back in Step 2 I have 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.
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Re: Trig Induction: show sin(2^n @) = 2^n sin(@)cos(@)cos(2@

Postby Nar on Sun Jan 05, 2014 10:31 pm

OK thanks very much, just 2 questions...

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

Is this an identity, I don't know?

FInal step


I know I can use index law for the first part


But I don't know what to do with the ..... at the end

Thanks very much, I think I'm missing a very basic concept here!
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Re: Trig Induction: show sin(2^n @) = 2^n sin(@)cos(@)cos(2@  TOPIC_SOLVED

Postby maggiemagnet on Mon Jan 06, 2014 2:41 am

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 ..... at the end

I plugged into the formula and then simplified.
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Re: Trig Induction: show sin(2^n @) = 2^n sin(@)cos(@)cos(2@

Postby Nar on Mon Jan 06, 2014 10:39 am

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



I know the at the front becomes but what happens with the both cos terms at the end?
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Postby stapel_eliz on Mon Jan 06, 2014 1:17 pm

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



...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?
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