Answered

At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

Mathematics III Sem 2
4.9.3 Quiz: Trigonometric Identities
Question 1 of 10

Which of the following are identities? Check all that apply.

A. [tex][tex]$\cos (x+y)+\cos (x-y)=2 \cos x \cos y$[/tex][/tex]
B. [tex][tex]$\sin (x+y)+\sin (x-y)=2 \cos x \sin y$[/tex][/tex]
C. [tex][tex]$\cos (x+y)-\cos (x-y)=2 \cos x \cos y$[/tex][/tex]
D. [tex][tex]$\tan (x-\pi)=\tan x$[/tex][/tex]

Sagot :

GinnyAnswer
Let's examine each of the given trigonometric identities to determine which are true.

### Identity A
[tex]\[ \cos (x+y) + \cos (x-y) = 2 \cos x \cos y \][/tex]

To verify this identity, we'll use the sum-to-product identities:
[tex]\[ \cos (x+y) + \cos (x-y) = 2 \cos \left(\frac{(x+y) + (x-y)}{2}\right) \cos \left(\frac{(x+y) - (x-y)}{2}\right) \][/tex]

Simplifying each argument:
[tex]\[ \frac{(x+y) + (x-y)}{2} = \frac{2x}{2} = x \][/tex]
[tex]\[ \frac{(x+y) - (x-y)}{2} = \frac{2y}{2} = y \][/tex]

So we have:
[tex]\[ \cos (x+y) + \cos (x-y) = 2 \cos x \cos y \][/tex]

Thus, identity A is true.

### Identity B
[tex]\[ \sin (x+y) + \sin (x-y) = 2 \cos x \sin y \][/tex]

To verify this identity, we'll use the sum-to-product identities:
[tex]\[ \sin (x+y) + \sin (x-y) = 2 \sin \left(\frac{(x+y) + (x-y)}{2}\right) \cos \left(\frac{(x+y) - (x-y)}{2}\right) \][/tex]

Simplifying each argument:
[tex]\[ \frac{(x+y) + (x-y)}{2} = \frac{2x}{2} = x \][/tex]
[tex]\[ \frac{(x+y) - (x-y)}{2} = \frac{2y}{2} = y \][/tex]

So we have:
[tex]\[ \sin (x+y) + \sin (x-y) = 2 \sin x \cos y \][/tex]

Given identity states [tex]\( 2 \cos x \sin y \)[/tex] instead of [tex]\( 2 \sin x \cos y \)[/tex], thus this identity is false.

### Identity C
[tex]\[ \cos (x+y) - \cos (x-y)=2 \sin x \sin y \][/tex]

To verify this identity, we'll use another sum-to-product identity:
[tex]\[ \cos (x+y) - \cos (x-y) = -2 \sin \left(\frac{(x+y) + (x-y)}{2}\right) \sin \left(\frac{(x+y) - (x-y)}{2}\right) \][/tex]

Simplifying each argument:
[tex]\[ \frac{(x+y) + (x-y)}{2} = \frac{2x}{2} = x \][/tex]
[tex]\[ \frac{(x+y) - (x-y)}{2} = \frac{2y}{2} = y \][/tex]

So we have:
[tex]\[ \cos (x+y) - \cos (x-y) = -2 \sin x \sin y \][/tex]

Given identity states [tex]\(2 \sin x \sin y \)[/tex] without the negative sign, so this identity is false.

### Identity D
[tex]\[ \tan (x - \pi) = \tan x \][/tex]

To verify this, we use the periodic property of the tangent function. The tangent function has a period of [tex]\(\pi\)[/tex], which means:
[tex]\[ \tan (x - \pi) = \tan x \][/tex]

This identity is true.

Therefore, the true identities are:
- A. [tex]\(\cos (x+y) + \cos (x-y) = 2 \cos x \cos y\)[/tex]
- D. [tex]\(\tan (x-\pi) = \tan x\)[/tex]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.