Answer :
To determine whether the given expressions are trigonometric identities, we need to simplify each expression step-by-step and check if they hold true for all values of \(x\).
### Expression A
[tex]\[ (\sin x + \cos x)^2 = 1 + \sin 2x \][/tex]
1. Expand the left-hand side:
[tex]\[ (\sin x + \cos x)^2 = \sin^2 x + 2 \sin x \cos x + \cos^2 x \][/tex]
2. Use the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\):
[tex]\[ 1 + 2 \sin x \cos x \][/tex]
3. Recognize that \(2 \sin x \cos x = \sin 2x\):
[tex]\[ 1 + \sin 2x \][/tex]
The left-hand side simplifies exactly to the right-hand side. Therefore, this is an identity.
### Expression B
[tex]\[ \frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \tan x \][/tex]
1. Use the sum-to-product identities for the numerator and denominator.
[tex]\[ \sin 3x - \sin x = 2 \cos \left(\frac{3x + x}{2}\right) \sin \left(\frac{3x - x}{2}\right) = 2 \cos(2x) \sin x \][/tex]
2. Similarly, for the denominator, use the sum-to-product identities:
[tex]\[ \cos 3x + \cos x = 2 \cos \left(\frac{3x + x}{2}\right) \cos \left(\frac{3x - x}{2}\right) = 2 \cos(2x) \cos x \][/tex]
3. Substituting these into the given expression:
[tex]\[ \frac{2 \cos(2x) \sin x}{2 \cos(2x) \cos x} = \frac{\sin x}{\cos x} = \tan x \][/tex]
This simplification confirms that the expression is an identity.
### Expression C
[tex]\[ \sin 6x = 2 \sin 3x \cos 3x \][/tex]
Using the double-angle formula \(\sin 2\theta = 2 \sin \theta \cos \theta\):
[tex]\[ \sin 6x = \sin (2 \cdot 3x) = 2 \sin 3x \cos 3x \][/tex]
This simplification confirms that the expression is an identity.
### Expression D
[tex]\[ \frac{\sin 3x}{\sin x \cos x} = 4 \cos x - \sec x \][/tex]
First, simplify the left-hand side:
[tex]\[ \frac{\sin 3x}{\sin x \cos x} \][/tex]
Using the triple angle identity for sine:
[tex]\[ \sin 3x = 3 \sin x - 4 \sin^3 x \][/tex]
Substitute back into the expression:
[tex]\[ \frac{3 \sin x - 4 \sin^3 x}{\sin x \cos x} = \frac{3 \sin x}{\sin x \cos x} - \frac{4 \sin^3 x}{\sin x \cos x} \][/tex]
[tex]\[ = \frac{3}{\cos x} - \frac{4 \sin^2 x}{\cos x} \][/tex]
Using \(\sin^2 x = 1 - \cos^2 x\):
[tex]\[ = \frac{3}{\cos x} - \frac{4 (1-\cos^2 x)}{\cos x} \][/tex]
[tex]\[ = \sec x - \frac{4 (1-\cos^2 x)}{\cos x} \][/tex]
[tex]\[ = \sec x - 4 \left(\frac{1}{\cos x} - \cos x \right) \][/tex]
[tex]\[ = \sec x - 4 \sec x + 4 \cos x \][/tex]
Simplify:
[tex]\[ \sec x - 4 \sec x + 4 \cos x = -3 \sec x + 4 \cos x \][/tex]
This does not match the right-hand side of the given expression \(4 \cos x - \sec x\).
Thus, the expressions that are identities are:
- Expression A
- Expression C
### Expression A
[tex]\[ (\sin x + \cos x)^2 = 1 + \sin 2x \][/tex]
1. Expand the left-hand side:
[tex]\[ (\sin x + \cos x)^2 = \sin^2 x + 2 \sin x \cos x + \cos^2 x \][/tex]
2. Use the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\):
[tex]\[ 1 + 2 \sin x \cos x \][/tex]
3. Recognize that \(2 \sin x \cos x = \sin 2x\):
[tex]\[ 1 + \sin 2x \][/tex]
The left-hand side simplifies exactly to the right-hand side. Therefore, this is an identity.
### Expression B
[tex]\[ \frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \tan x \][/tex]
1. Use the sum-to-product identities for the numerator and denominator.
[tex]\[ \sin 3x - \sin x = 2 \cos \left(\frac{3x + x}{2}\right) \sin \left(\frac{3x - x}{2}\right) = 2 \cos(2x) \sin x \][/tex]
2. Similarly, for the denominator, use the sum-to-product identities:
[tex]\[ \cos 3x + \cos x = 2 \cos \left(\frac{3x + x}{2}\right) \cos \left(\frac{3x - x}{2}\right) = 2 \cos(2x) \cos x \][/tex]
3. Substituting these into the given expression:
[tex]\[ \frac{2 \cos(2x) \sin x}{2 \cos(2x) \cos x} = \frac{\sin x}{\cos x} = \tan x \][/tex]
This simplification confirms that the expression is an identity.
### Expression C
[tex]\[ \sin 6x = 2 \sin 3x \cos 3x \][/tex]
Using the double-angle formula \(\sin 2\theta = 2 \sin \theta \cos \theta\):
[tex]\[ \sin 6x = \sin (2 \cdot 3x) = 2 \sin 3x \cos 3x \][/tex]
This simplification confirms that the expression is an identity.
### Expression D
[tex]\[ \frac{\sin 3x}{\sin x \cos x} = 4 \cos x - \sec x \][/tex]
First, simplify the left-hand side:
[tex]\[ \frac{\sin 3x}{\sin x \cos x} \][/tex]
Using the triple angle identity for sine:
[tex]\[ \sin 3x = 3 \sin x - 4 \sin^3 x \][/tex]
Substitute back into the expression:
[tex]\[ \frac{3 \sin x - 4 \sin^3 x}{\sin x \cos x} = \frac{3 \sin x}{\sin x \cos x} - \frac{4 \sin^3 x}{\sin x \cos x} \][/tex]
[tex]\[ = \frac{3}{\cos x} - \frac{4 \sin^2 x}{\cos x} \][/tex]
Using \(\sin^2 x = 1 - \cos^2 x\):
[tex]\[ = \frac{3}{\cos x} - \frac{4 (1-\cos^2 x)}{\cos x} \][/tex]
[tex]\[ = \sec x - \frac{4 (1-\cos^2 x)}{\cos x} \][/tex]
[tex]\[ = \sec x - 4 \left(\frac{1}{\cos x} - \cos x \right) \][/tex]
[tex]\[ = \sec x - 4 \sec x + 4 \cos x \][/tex]
Simplify:
[tex]\[ \sec x - 4 \sec x + 4 \cos x = -3 \sec x + 4 \cos x \][/tex]
This does not match the right-hand side of the given expression \(4 \cos x - \sec x\).
Thus, the expressions that are identities are:
- Expression A
- Expression C
