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

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The identity 1 + cos(x) + cos(2x) = 1/2 + (sin(5/2x) / (2sin(1/2x))) is verified through a series of trigonometric transformations. Starting from the left side, it is rewritten using known identities, leading to the expression involving cosines and sines. The right side is manipulated using the triple-angle identity for sine and product-to-sum identities, ultimately simplifying to match the left side. The verification process demonstrates the equality by systematically applying trigonometric identities and algebraic properties. The identity is confirmed as shown.
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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.
 
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Hello Catlover0330,

We are given to verify:

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

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

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

Using the identities:

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

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

We may write:

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

Factor the right side:

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

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

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

to obtain:

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

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

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

to obtain:

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

Multiply the right side by:

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

to obtain:

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

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

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

to obtain:

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

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

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

to obtain:

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

Rewrite the right side using the algebraic property:

$$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$$

to obtain:

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

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

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

Shown as desired.
 
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