Answer :
To prove the given trigonometric identity:
[tex]\[ \cos \left(36^{\circ}-\theta\right) \cdot \cos \left(36^{\circ}+\theta\right) + \cos \left(54^{\circ}+\theta\right) \cdot \cos \left(54^{\circ}-\theta\right) = \cos 2\theta \][/tex]
we can use trigonometric product-to-sum identities.
### Step 1: Using Product-to-Sum Identities
The product-to-sum identities for cosine are given by:
[tex]\[ \cos A \cos B = \frac{1}{2} \left[ \cos (A + B) + \cos (A - B) \right] \][/tex]
Let's apply this identity to each product term in the given expression.
#### Term 1: [tex]\(\cos (36^\circ - \theta) \cos (36^\circ + \theta)\)[/tex]
Applying the product-to-sum identity:
[tex]\[ \cos (36^\circ - \theta) \cos (36^\circ + \theta) = \frac{1}{2} \left[ \cos \left( (36^\circ - \theta) + (36^\circ + \theta) \right) + \cos \left( (36^\circ - \theta) - (36^\circ + \theta) \right) \right] \][/tex]
Simplifying the cosine arguments:
[tex]\[ \cos (36^\circ - \theta) \cos (36^\circ + \theta) = \frac{1}{2} \left[ \cos (72^\circ) + \cos (-2\theta) \right] \][/tex]
Since [tex]\(\cos (-2\theta) = \cos 2\theta\)[/tex]:
[tex]\[ \cos (36^\circ - \theta) \cos (36^\circ + \theta) = \frac{1}{2} \left[ \cos (72^\circ) + \cos 2\theta \right] \][/tex]
#### Term 2: [tex]\(\cos (54^\circ + \theta) \cos (54^\circ - \theta)\)[/tex]
Applying the product-to-sum identity here as well:
[tex]\[ \cos (54^\circ + \theta) \cos (54^\circ - \theta) = \frac{1}{2} \left[ \cos \left( (54^\circ + \theta) + (54^\circ - \theta) \right) + \cos \left( (54^\circ + \theta) - (54^\circ - \theta) \right) \right] \][/tex]
Simplifying the cosine arguments:
[tex]\[ \cos (54^\circ + \theta) \cos (54^\circ - \theta) = \frac{1}{2} \left[ \cos (108^\circ) + \cos (2\theta) \right] \][/tex]
### Step 2: Adding the Two Terms
Now, let's add the two simplified terms together:
[tex]\[ \cos (36^\circ - \theta) \cos (36^\circ + \theta) + \cos (54^\circ + \theta) \cos (54^\circ - \theta) \][/tex]
This becomes:
[tex]\[ \frac{1}{2} \left[ \cos (72^\circ) + \cos 2\theta \right] + \frac{1}{2} \left[ \cos (108^\circ) + \cos 2\theta \right] \][/tex]
We can combine these terms:
[tex]\[ \frac{1}{2} \left[ \cos (72^\circ) + \cos (108^\circ) + \cos 2\theta + \cos 2\theta \right] \][/tex]
This simplifies to:
[tex]\[ \frac{1}{2} \left[ \cos (72^\circ) + \cos (108^\circ) + 2\cos 2\theta \right] \][/tex]
### Step 3: Evaluating Cosine Values
Next, we need to know the values of [tex]\(\cos(72^\circ)\)[/tex] and [tex]\(\cos(108^\circ)\)[/tex].
Recall that:
[tex]\[ \cos(108^\circ) = -\cos(72^\circ) \][/tex]
Therefore:
[tex]\[ \cos (72^\circ) + \cos (108^\circ) = \cos (72^\circ) - \cos (72^\circ) = 0 \][/tex]
### Step 4: Substituting Back
Substitute [tex]\(0\)[/tex] for [tex]\(\cos(72^\circ) + \cos(108^\circ)\)[/tex]:
[tex]\[ \frac{1}{2} \left[ 0 + 2\cos 2\theta \right] \][/tex]
This simplifies to:
[tex]\[ \frac{1}{2} \times 2 \cos 2 \theta = \cos 2 \theta \][/tex]
Thus, the given expression is indeed equal to [tex]\(\cos 2\theta\)[/tex], proving the identity:
[tex]\[ \cos (36^\circ - \theta) \cos (36^\circ + \theta) + \cos (54^\circ + \theta) \cos (54^\circ - \theta) = \cos 2 \theta \][/tex]
[tex]\[ \cos \left(36^{\circ}-\theta\right) \cdot \cos \left(36^{\circ}+\theta\right) + \cos \left(54^{\circ}+\theta\right) \cdot \cos \left(54^{\circ}-\theta\right) = \cos 2\theta \][/tex]
we can use trigonometric product-to-sum identities.
### Step 1: Using Product-to-Sum Identities
The product-to-sum identities for cosine are given by:
[tex]\[ \cos A \cos B = \frac{1}{2} \left[ \cos (A + B) + \cos (A - B) \right] \][/tex]
Let's apply this identity to each product term in the given expression.
#### Term 1: [tex]\(\cos (36^\circ - \theta) \cos (36^\circ + \theta)\)[/tex]
Applying the product-to-sum identity:
[tex]\[ \cos (36^\circ - \theta) \cos (36^\circ + \theta) = \frac{1}{2} \left[ \cos \left( (36^\circ - \theta) + (36^\circ + \theta) \right) + \cos \left( (36^\circ - \theta) - (36^\circ + \theta) \right) \right] \][/tex]
Simplifying the cosine arguments:
[tex]\[ \cos (36^\circ - \theta) \cos (36^\circ + \theta) = \frac{1}{2} \left[ \cos (72^\circ) + \cos (-2\theta) \right] \][/tex]
Since [tex]\(\cos (-2\theta) = \cos 2\theta\)[/tex]:
[tex]\[ \cos (36^\circ - \theta) \cos (36^\circ + \theta) = \frac{1}{2} \left[ \cos (72^\circ) + \cos 2\theta \right] \][/tex]
#### Term 2: [tex]\(\cos (54^\circ + \theta) \cos (54^\circ - \theta)\)[/tex]
Applying the product-to-sum identity here as well:
[tex]\[ \cos (54^\circ + \theta) \cos (54^\circ - \theta) = \frac{1}{2} \left[ \cos \left( (54^\circ + \theta) + (54^\circ - \theta) \right) + \cos \left( (54^\circ + \theta) - (54^\circ - \theta) \right) \right] \][/tex]
Simplifying the cosine arguments:
[tex]\[ \cos (54^\circ + \theta) \cos (54^\circ - \theta) = \frac{1}{2} \left[ \cos (108^\circ) + \cos (2\theta) \right] \][/tex]
### Step 2: Adding the Two Terms
Now, let's add the two simplified terms together:
[tex]\[ \cos (36^\circ - \theta) \cos (36^\circ + \theta) + \cos (54^\circ + \theta) \cos (54^\circ - \theta) \][/tex]
This becomes:
[tex]\[ \frac{1}{2} \left[ \cos (72^\circ) + \cos 2\theta \right] + \frac{1}{2} \left[ \cos (108^\circ) + \cos 2\theta \right] \][/tex]
We can combine these terms:
[tex]\[ \frac{1}{2} \left[ \cos (72^\circ) + \cos (108^\circ) + \cos 2\theta + \cos 2\theta \right] \][/tex]
This simplifies to:
[tex]\[ \frac{1}{2} \left[ \cos (72^\circ) + \cos (108^\circ) + 2\cos 2\theta \right] \][/tex]
### Step 3: Evaluating Cosine Values
Next, we need to know the values of [tex]\(\cos(72^\circ)\)[/tex] and [tex]\(\cos(108^\circ)\)[/tex].
Recall that:
[tex]\[ \cos(108^\circ) = -\cos(72^\circ) \][/tex]
Therefore:
[tex]\[ \cos (72^\circ) + \cos (108^\circ) = \cos (72^\circ) - \cos (72^\circ) = 0 \][/tex]
### Step 4: Substituting Back
Substitute [tex]\(0\)[/tex] for [tex]\(\cos(72^\circ) + \cos(108^\circ)\)[/tex]:
[tex]\[ \frac{1}{2} \left[ 0 + 2\cos 2\theta \right] \][/tex]
This simplifies to:
[tex]\[ \frac{1}{2} \times 2 \cos 2 \theta = \cos 2 \theta \][/tex]
Thus, the given expression is indeed equal to [tex]\(\cos 2\theta\)[/tex], proving the identity:
[tex]\[ \cos (36^\circ - \theta) \cos (36^\circ + \theta) + \cos (54^\circ + \theta) \cos (54^\circ - \theta) = \cos 2 \theta \][/tex]
