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]



Answer :

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]