Answered

Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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 revisit us for more reliable answers to any questions you may have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.