Answer :
Sure, let’s break down the proof step by step.
Given the equation to prove:
[tex]\[ 1 + \cos^2(2\theta) = 2(\cos^4(\theta) + \sin^4(\theta)) \][/tex]
### Step 1: Express \(\cos^2(2\theta)\) using trigonometric identities.
Recall the double-angle identity for cosine:
[tex]\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \][/tex]
Squaring both sides, we get:
[tex]\[ \cos^2(2\theta) = (2\cos^2(\theta) - 1)^2 \][/tex]
[tex]\[ \cos^2(2\theta) = 4\cos^4(\theta) - 4\cos^2(\theta) + 1 \][/tex]
### Step 2: Substitute \(\cos^2(2\theta)\) back into the left-hand side (LHS).
The LHS given is:
[tex]\[ 1 + \cos^2(2\theta) \][/tex]
Substitute the expression for \(\cos^2(2\theta)\):
[tex]\[ 1 + \cos^2(2\theta) = 1 + 4\cos^4(\theta) - 4\cos^2(\theta) + 1 \][/tex]
[tex]\[ 1 + 4\cos^4(\theta) - 4\cos^2(\theta) + 1 = 4\cos^4(\theta) - 4\cos^2(\theta) + 2 \][/tex]
### Step 3: Simplify the right-hand side (RHS).
The RHS given is:
[tex]\[ 2(\cos^4(\theta) + \sin^4(\theta)) \][/tex]
We can use the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\) to rewrite the expressions involving \(\sin^4(\theta)\):
Notice that:
[tex]\[ \sin^4(\theta) = (\sin^2(\theta))^2 \][/tex]
[tex]\[ \cos^4(\theta) = (\cos^2(\theta))^2 \][/tex]
Thus:
[tex]\[ 2(\cos^4(\theta) + \sin^4(\theta)) = 2((\cos^2(\theta))^2 + (\sin^2(\theta))^2) \][/tex]
Break it into simpler terms:
[tex]\[ 2(\cos^4(\theta) + \sin^4(\theta)) = 2(\cos^4(\theta) + (1 - \cos^2(\theta))^2) \][/tex]
[tex]\[ = 2(\cos^4(\theta) + 1 - 2\cos^2(\theta) + \cos^4(\theta)) \][/tex]
[tex]\[ = 2(2\cos^4(\theta) - 2\cos^2(\theta) + 1) \][/tex]
[tex]\[ = 4\cos^4(\theta) - 4\cos^2(\theta) + 2 \][/tex]
### Conclusion: Compare LHS with RHS.
Notice that both the simplified LHS and RHS give us the same expression:
[tex]\[ 4\cos^4(\theta) - 4\cos^2(\theta) + 2 \][/tex]
Thus, we have shown that:
[tex]\[ 1 + \cos^2(2\theta) = 2(\cos^4(\theta) + \sin^4(\theta)) \][/tex]
Therefore, the proof is complete. The given equation is true.
Given the equation to prove:
[tex]\[ 1 + \cos^2(2\theta) = 2(\cos^4(\theta) + \sin^4(\theta)) \][/tex]
### Step 1: Express \(\cos^2(2\theta)\) using trigonometric identities.
Recall the double-angle identity for cosine:
[tex]\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \][/tex]
Squaring both sides, we get:
[tex]\[ \cos^2(2\theta) = (2\cos^2(\theta) - 1)^2 \][/tex]
[tex]\[ \cos^2(2\theta) = 4\cos^4(\theta) - 4\cos^2(\theta) + 1 \][/tex]
### Step 2: Substitute \(\cos^2(2\theta)\) back into the left-hand side (LHS).
The LHS given is:
[tex]\[ 1 + \cos^2(2\theta) \][/tex]
Substitute the expression for \(\cos^2(2\theta)\):
[tex]\[ 1 + \cos^2(2\theta) = 1 + 4\cos^4(\theta) - 4\cos^2(\theta) + 1 \][/tex]
[tex]\[ 1 + 4\cos^4(\theta) - 4\cos^2(\theta) + 1 = 4\cos^4(\theta) - 4\cos^2(\theta) + 2 \][/tex]
### Step 3: Simplify the right-hand side (RHS).
The RHS given is:
[tex]\[ 2(\cos^4(\theta) + \sin^4(\theta)) \][/tex]
We can use the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\) to rewrite the expressions involving \(\sin^4(\theta)\):
Notice that:
[tex]\[ \sin^4(\theta) = (\sin^2(\theta))^2 \][/tex]
[tex]\[ \cos^4(\theta) = (\cos^2(\theta))^2 \][/tex]
Thus:
[tex]\[ 2(\cos^4(\theta) + \sin^4(\theta)) = 2((\cos^2(\theta))^2 + (\sin^2(\theta))^2) \][/tex]
Break it into simpler terms:
[tex]\[ 2(\cos^4(\theta) + \sin^4(\theta)) = 2(\cos^4(\theta) + (1 - \cos^2(\theta))^2) \][/tex]
[tex]\[ = 2(\cos^4(\theta) + 1 - 2\cos^2(\theta) + \cos^4(\theta)) \][/tex]
[tex]\[ = 2(2\cos^4(\theta) - 2\cos^2(\theta) + 1) \][/tex]
[tex]\[ = 4\cos^4(\theta) - 4\cos^2(\theta) + 2 \][/tex]
### Conclusion: Compare LHS with RHS.
Notice that both the simplified LHS and RHS give us the same expression:
[tex]\[ 4\cos^4(\theta) - 4\cos^2(\theta) + 2 \][/tex]
Thus, we have shown that:
[tex]\[ 1 + \cos^2(2\theta) = 2(\cos^4(\theta) + \sin^4(\theta)) \][/tex]
Therefore, the proof is complete. The given equation is true.
