1. ## trignometric problem

1. LHS=2cos 2^a+1/2cos a+1= RHS=(2cos a-1) (2cos 2a-1) (2cos 2^2a-1).....(2cos 2^n-1 a-1)

here a is the angle n is number

we have to prove LHS=RHS

Hello prasum
Originally Posted by prasum
1.2cos 2^a+1/2cos a+1=(2cos a-1) (2cos 2a-1) (2cos 2^2a-1).....(2cos 2^n-1 a-1)

I think the reason you have not had a reply is that no-one understands what you have written. You appear to have written:
$1.2\cos2^a+\tfrac12\cos a+1=(2\cos a-1) (2\cos 2a-1) (2\cos 2^2a-1).....(2\cos 2^{n-1} a-1)$
So, two questions:
What did you mean to write?

So what? Do we have to prove that the two sides are equal?

3. Thats not 1/2 thats whole first part is divided

4. Hi Prasum

Use Latex or if you dont know it and than try to use your brain your question is still not at all clear...

Here is another try is this correct
$
\frac{2\cos(2^{a})+1}{2\cos(a)+1}=(2\cos a-1) (2\cos 2a-1) (2\cos 2^2a-1).....(2\cos 2^{n-1} a-1)$

5. it is 2cos 2^n a+1 what yu have written is correct

6. Hello prasum

If you're serious about wanting help on this Forum - and, indeed, serious about improving your mathematical skills - you must learn to be more accurate in writing down mathematical expressions. This will save everyone a good deal of time, and help you to acquire the skills you need to solve problems like this.

I think the question is this:
Prove, for any positive integer $n$, and any value of $a$, that
$\frac{2\cos2^na+1}{2\cos a +1}=(2\cos a-1)(2\cos 2a-1)(2\cos 2^2a-1)...(2\cos 2^{n-1}a-1)$
This is done by Mathematical Induction, as follows.

Let $P(n)$ be the propositional function:
$(2\cos a-1)(2\cos 2a-1)(2\cos 2^2a-1)...(2\cos 2^{n-1}a-1)=\frac{2\cos2^na+1}{2\cos a +1}$
(For convenience, I've swapped the sides of the equation.)

Then, multiplying both sides of the equation by $(2\cos2^na-1)$:
$P(n)\Rightarrow (2\cos a-1)(2\cos 2a-1)(2\cos 2^2a-1)...(2\cos 2^{n-1}a-1)(2\cos2^na-1)$ $=\frac{(2\cos2^na+1)(2\cos2^na-1)}{2\cos a +1}$
$=\frac{4\cos^22^na-1}{2\cos a + 1}$, using $(a+b)(a-b) = a^2-b^2$

$=\frac{2(2\cos^22^na-1)+1}{2\cos a + 1}$

$=\frac{2\cos2^{n+1}a+1}{2\cos a + 1}$, using $\cos 2\theta = 2\cos^2\theta-1$

$\Rightarrow P(n+1)$
Now $P(1)$ states:
$2\cos a -1=\frac{2\cos2a+1}{2\cos a +1}$
The RHS is:
$\frac{2(2\cos^2a-1)+1}{2\cos a +1}$
$=\frac{4\cos^2a-1}{2\cos a +1}$

$=\frac{(2\cos a+1)(2\cos a -1)}{2\cos a +1}$

$=2\cos a -1$

So $P(1)$ is true. Therefore, by Induction, $P(n)$ is true for all positive integers, $n$.