Verify Trig Identity: 1+cosx+cos2x=1/2+(sin5/2x)/(2sin1/2x) - Catlover0330

[MarkFL]

Here is the question:

Verify the following identity: 1 + cosx + cos2x = 1/2 + (sin5/2x) / (2sin1/2x)?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Catlover0330,

We are given to verify:

\(\displaystyle 1+\cos(x)+\cos(2x)=\frac{1}{2}+ \frac{\sin\left(\frac{5}{2}x \right)}{2 \sin\left(\frac{1}{2}x \right)}\)

Let's begin with the left side of the identity and rewrite it as follows:

\(\displaystyle 1+\cos(x)+\cos(2x)=1+3+4\cos(x)+\cos(2x)-3-3\cos(x)\)

Using the identities:

\(\displaystyle 8\cos^4(\theta)=3+4\cos(2\theta)+\cos(4\theta)\)

\(\displaystyle 6\cos^2(\theta)=3+3\cos(2\theta)\)

We may write:

\(\displaystyle 1+\cos(x)+\cos(2x)=1+8\cos^4\left(\frac{1}{2}x \right)-6\cos^2\left(\frac{1}{2}x \right)\)

Factor the right side:

\(\displaystyle 1+\cos(x)+\cos(2x)=\left(4\cos^2\left(\frac{1}{2}x \right)-1 \right)\left(2\cos^2\left(\frac{1}{2}x \right)-1 \right)\)

Rewrite the first factor and use the double-angle identity for cosine on the second factor:

\(\displaystyle \cos(2\theta)=2\cos^2(\theta)-1\)

to obtain:

\(\displaystyle 1+\cos(x)+\cos(2x)=\left(3-4\left(1-\cos^2\left(\frac{1}{2}x \right) \right) \right)\cos(x)\)

To the first factor on the right, apply the Pythagorean identity:

\(\displaystyle \sin^2(\theta)=1-\cos^2(\theta)\)

to obtain:

\(\displaystyle 1+\cos(x)+\cos(2x)=\left(3-4\sin^2\left(\frac{1}{2}x \right) \right)\cos(x)\)

Multiply the right side by:

\(\displaystyle 1=\frac{\sin\left(\frac{1}{2}x \right)}{\sin\left(\frac{1}{2}x \right)}\)

to obtain:

\(\displaystyle 1+\cos(x)+\cos(2x)=\frac{\left(3\sin\left(\frac{1}{2}x \right)-4\sin^3\left(\frac{1}{2}x \right) \right)\cos(x)}{\sin\left(\frac{1}{2}x \right)}\)

To the first factor in the numerator on the right, apply the triple-angle identity for sine:

\(\displaystyle \sin(3\theta)=3\sin(\theta)-4\sin^3(\theta)\)

to obtain:

\(\displaystyle 1+\cos(x)+\cos(2x)=\frac{\sin\left(\frac{3}{2}x \right)\cos(x)}{\sin\left(\frac{1}{2}x \right)}\)

To the numerator on the right, apply the product-to-sum identity:

\(\displaystyle \sin(\alpha)\cos(\beta)=\frac{\sin(\alpha-\beta)+\sin(\alpha+\beta)}{2}\)

to obtain:

\(\displaystyle 1+\cos(x)+\cos(2x)=\frac{\sin\left(\frac{1}{2}x \right)+\sin\left(\frac{5}{2}x \right)}{2\sin\left(\frac{1}{2}x \right)}\)

Rewrite the right side using the algebraic property:

\(\displaystyle \frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}\)

to obtain:

\(\displaystyle 1+\cos(x)+\cos(2x)=\frac{\sin\left(\frac{1}{2}x \right)}{2\sin\left(\frac{1}{2}x \right)}+\frac{\sin\left(\frac{5}{2}x \right)}{2\sin\left(\frac{1}{2}x \right)}\)

Divide out common factor in numerator and denominator of first term on the right:

\(\displaystyle 1+\cos(x)+\cos(2x)=\frac{1}{2}+\frac{\sin\left( \frac{5}{2}x \right)}{2\sin\left(\frac{1}{2}x \right)}\)

Shown as desired.