Please help me make this true

[Karn00k]

Please help

1+sina=2cos((n/4)/(a/2))
a-alpha
n-pi
I don't understand what I could do with either side

 

[DrPhil]

1+sina=2cos((n/4)/(a/2))
a-alpha
n-pi
I don't understand what I could do with either side

You have used enough parentheses to make the expression on the right completely unambiguous, but it must be wrong. You wind up with alpha in the denominator, and there are NO identities for reciprocal angles. The problem would be sensible if the division symbol is either a + or \(\displaystyle -\). In that case you can operate on the right side by using the cos(A±B) identity, and then the half-angle formulas. And always be ready to use the most fundamental identity of all: sin^2(a) + cos^2(a) = 1.

Please check your typing, and then show us how far you can get.

 

[Karn00k]

You have used enough parentheses to make the expression on the right completely unambiguous, but it must be wrong. You wind up with alpha in the denominator, and there are NO identities for reciprocal angles. The problem would be sensible if the division symbol is either a + or \(\displaystyle -\). In that case you can operate on the right side by using the cos(A±B) identity, and then the half-angle formulas. And always be ready to use the most fundamental identity of all: sin^2(a) + cos^2(a) = 1.

Please check your typing, and then show us how far you can get.

Here it is:

I would show you my work if I had any. Unfortunately, I still don't understand what to do with the right side. I think that I should work on the right side to get the left side because there is absolutely nothing that you can do with the left side.
So, could you please help me get started on this problem?

 

[MarkFL]

I would begin, as is somewhat traditional, with the left side, and using the aforementioned Pythagorean identity along with the double-angle identity for sine to write it as:

\(\displaystyle \cos^2\left(\frac{a}{2} \right)+2\sin\left(\frac{a}{2} \right)\cos\left(\frac{a}{2} \right)+\sin^2\left(\frac{a}{2} \right)\)

Now, observing that this is the square of a binomial, we may write this as:

\(\displaystyle \left(\cos\left(\frac{a}{2} \right)+\sin\left(\frac{a}{2} \right) \right)^2\)

Now, seeing that \(\displaystyle \sin\left(\frac{\pi}{4} \right)=\cos\left(\frac{\pi}{4} \right)=\dfrac{1}{\sqrt{2}}\) we may write this as:

\(\displaystyle \left(\sqrt{2}\left(\cos\left(\frac{\pi}{4} \right)\cos\left(\frac{a}{2} \right)+\sin\left(\frac{\pi}{4} \right)\sin\left(\frac{a}{2} \right) \right) \right)^2\)

Can you finish?