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Sagot :
To determine which of the given equations are identities, we will transform and simplify both sides of each equation to see if they are indeed equal. Let's go through them one by one.
### A. [tex]\(\frac{\sin x + \sin 5x}{\cos x + \cos 5x} = \tan 3x\)[/tex]
Left Side:
We will use the sum-to-product identities for sine and cosine:
[tex]\[ \sin a + \sin b = 2 \sin\left(\frac{a + b}{2}\right) \cos\left(\frac{a - b}{2}\right) \][/tex]
[tex]\[ \cos a + \cos b = 2 \cos\left(\frac{a + b}{2}\right) \cos\left(\frac{a - b}{2}\right) \][/tex]
Applying these to [tex]\(\sin x + \sin 5x\)[/tex]:
[tex]\[ \sin x + \sin 5x = 2 \sin\left(\frac{6x}{2}\right) \cos\left(\frac{-4x}{2}\right) = 2 \sin(3x) \cos(2x) \][/tex]
Now for [tex]\(\cos x + \cos 5x\)[/tex]:
[tex]\[ \cos x + \cos 5x = 2 \cos\left(\frac{6x}{2}\right) \cos\left(\frac{4x}{2}\right) = 2 \cos(3x) \cos(2x) \][/tex]
So the left side becomes:
[tex]\[ \frac{\sin x + \sin 5x}{\cos x + \cos 5x} = \frac{2 \sin(3x) \cos(2x)}{2 \cos(3x) \cos(2x)} = \frac{\sin(3x)}{\cos(3x)} = \tan(3x) \][/tex]
Thus, [tex]\(\frac{\sin x + \sin 5x}{\cos x + \cos 5x} = \tan 3x\)[/tex] is indeed an identity.
### B. [tex]\((\sin x - \cos x)^2 = 1 + \sin 2x\)[/tex]
Expanding the left side:
[tex]\[ (\sin x - \cos x)^2 = \sin^2 x - 2 \sin x \cos x + \cos^2 x \][/tex]
Using the Pythagorean identity, [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex]:
[tex]\[ \sin^2 x - 2 \sin x \cos x + \cos^2 x = 1 - 2 \sin x \cos x \][/tex]
Using the double angle identity, [tex]\(\sin 2x = 2 \sin x \cos x\)[/tex]:
[tex]\[ 1 - 2 \sin x \cos x = 1 - \sin 2x \][/tex]
So, [tex]\((\sin x - \cos x)^2 = 1 - \sin 2x\)[/tex]. This does not equal [tex]\(1 + \sin 2x\)[/tex], so this is not an identity.
### C. [tex]\(\sin 8x = 2 \sin 4x \cos 4x\)[/tex]
Using the double angle identity, [tex]\(\sin(2a) = 2 \sin a \cos a\)[/tex]:
[tex]\[ \sin 8x = \sin(2 \cdot 4x) = 2 \sin 4x \cos 4x \][/tex]
Thus, [tex]\(\sin 8x = 2 \sin 4x \cos 4x\)[/tex] is indeed an identity.
### D. [tex]\(\frac{1 - \tan^2 x}{2 \tan x} = \frac{1}{\tan 2x}\)[/tex]
First, let's use the double angle identity for tangent, [tex]\(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\)[/tex]:
[tex]\[ \frac{1}{\tan 2x} = \frac{1}{\frac{2 \tan x}{1 - \tan^2 x}} = \frac{1 - \tan^2 x}{2 \tan x} \][/tex]
Thus, [tex]\(\frac{1 - \tan^2 x}{2 \tan x} = \frac{1}{\tan 2x}\)[/tex] is indeed an identity.
### Conclusion
The identities are:
- A [tex]\(\frac{\sin x + \sin 5x}{\cos x + \cos 5x} = \tan 3x\)[/tex]
- C [tex]\(\sin 8x = 2 \sin 4x \cos 4x\)[/tex]
- D [tex]\(\frac{1 - \tan^2 x}{2 \tan x} = \frac{1}{\tan 2x}\)[/tex]
### Summary
- A: Identity
- B: Not an identity
- C: Identity
- D: Identity
### A. [tex]\(\frac{\sin x + \sin 5x}{\cos x + \cos 5x} = \tan 3x\)[/tex]
Left Side:
We will use the sum-to-product identities for sine and cosine:
[tex]\[ \sin a + \sin b = 2 \sin\left(\frac{a + b}{2}\right) \cos\left(\frac{a - b}{2}\right) \][/tex]
[tex]\[ \cos a + \cos b = 2 \cos\left(\frac{a + b}{2}\right) \cos\left(\frac{a - b}{2}\right) \][/tex]
Applying these to [tex]\(\sin x + \sin 5x\)[/tex]:
[tex]\[ \sin x + \sin 5x = 2 \sin\left(\frac{6x}{2}\right) \cos\left(\frac{-4x}{2}\right) = 2 \sin(3x) \cos(2x) \][/tex]
Now for [tex]\(\cos x + \cos 5x\)[/tex]:
[tex]\[ \cos x + \cos 5x = 2 \cos\left(\frac{6x}{2}\right) \cos\left(\frac{4x}{2}\right) = 2 \cos(3x) \cos(2x) \][/tex]
So the left side becomes:
[tex]\[ \frac{\sin x + \sin 5x}{\cos x + \cos 5x} = \frac{2 \sin(3x) \cos(2x)}{2 \cos(3x) \cos(2x)} = \frac{\sin(3x)}{\cos(3x)} = \tan(3x) \][/tex]
Thus, [tex]\(\frac{\sin x + \sin 5x}{\cos x + \cos 5x} = \tan 3x\)[/tex] is indeed an identity.
### B. [tex]\((\sin x - \cos x)^2 = 1 + \sin 2x\)[/tex]
Expanding the left side:
[tex]\[ (\sin x - \cos x)^2 = \sin^2 x - 2 \sin x \cos x + \cos^2 x \][/tex]
Using the Pythagorean identity, [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex]:
[tex]\[ \sin^2 x - 2 \sin x \cos x + \cos^2 x = 1 - 2 \sin x \cos x \][/tex]
Using the double angle identity, [tex]\(\sin 2x = 2 \sin x \cos x\)[/tex]:
[tex]\[ 1 - 2 \sin x \cos x = 1 - \sin 2x \][/tex]
So, [tex]\((\sin x - \cos x)^2 = 1 - \sin 2x\)[/tex]. This does not equal [tex]\(1 + \sin 2x\)[/tex], so this is not an identity.
### C. [tex]\(\sin 8x = 2 \sin 4x \cos 4x\)[/tex]
Using the double angle identity, [tex]\(\sin(2a) = 2 \sin a \cos a\)[/tex]:
[tex]\[ \sin 8x = \sin(2 \cdot 4x) = 2 \sin 4x \cos 4x \][/tex]
Thus, [tex]\(\sin 8x = 2 \sin 4x \cos 4x\)[/tex] is indeed an identity.
### D. [tex]\(\frac{1 - \tan^2 x}{2 \tan x} = \frac{1}{\tan 2x}\)[/tex]
First, let's use the double angle identity for tangent, [tex]\(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\)[/tex]:
[tex]\[ \frac{1}{\tan 2x} = \frac{1}{\frac{2 \tan x}{1 - \tan^2 x}} = \frac{1 - \tan^2 x}{2 \tan x} \][/tex]
Thus, [tex]\(\frac{1 - \tan^2 x}{2 \tan x} = \frac{1}{\tan 2x}\)[/tex] is indeed an identity.
### Conclusion
The identities are:
- A [tex]\(\frac{\sin x + \sin 5x}{\cos x + \cos 5x} = \tan 3x\)[/tex]
- C [tex]\(\sin 8x = 2 \sin 4x \cos 4x\)[/tex]
- D [tex]\(\frac{1 - \tan^2 x}{2 \tan x} = \frac{1}{\tan 2x}\)[/tex]
### Summary
- A: Identity
- B: Not an identity
- C: Identity
- D: Identity
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