Certainly! Here’s a detailed, step-by-step solution for verifying the given identity:
Given identity to verify:
[tex]\[ \sin^4(B) + \cos^4(B) = 1 - \frac{1}{2} \sin^2(2B) \][/tex]
### Step-by-Step Solution
1. Start with the Right-Hand Side (RHS) of the Identity:
The right-hand side of the given identity is:
[tex]\[ 1 - \frac{1}{2} \sin^2(2B) \][/tex]
2. Recall a Trigonometric Identity for [tex]\(\sin(2B)\)[/tex]:
We know from trigonometry that:
[tex]\[ \sin(2B) = 2 \sin(B) \cos(B) \][/tex]
3. Calculate [tex]\(\sin^2(2B)\)[/tex]:
From the identity for [tex]\(\sin(2B)\)[/tex], we can square both sides:
[tex]\[ \sin^2(2B) = (2 \sin(B) \cos(B))^2 = 4 \sin^2(B) \cos^2(B) \][/tex]
4. Substitute [tex]\(\sin^2(2B)\)[/tex] into the RHS:
Using the value of [tex]\(\sin^2(2B)\)[/tex]:
[tex]\[ 1 - \frac{1}{2} \sin^2(2B) = 1 - \frac{1}{2} (4 \sin^2(B) \cos^2(B)) \][/tex]
Simplify this expression:
[tex]\[ 1 - 2 \sin^2(B) \cos^2(B) \][/tex]
5. Compare the Left-Hand Side (LHS) with Simplified RHS:
Now let’s consider the left-hand side of the given identity:
[tex]\[ \sin^4(B) + \cos^4(B) \][/tex]
6. Rewrite [tex]\(\sin^4(B) + \cos^4(B)\)[/tex] in Terms of [tex]\(\sin^2(B)\)[/tex] and [tex]\(\cos^2(B)\)[/tex]:
Notice the algebraic identity:
[tex]\[ x^2 + y^2 = (x + y)^2 - 2xy \][/tex]
Applying this to [tex]\(\sin^2(B)\)[/tex] and [tex]\(\cos^2(B)\)[/tex], we get:
[tex]\[ \sin^4(B) + \cos^4(B) = (\sin^2(B))^2 + (\cos^2(B))^2 \][/tex]
We can rewrite this as:
[tex]\[ \sin^4(B) + \cos^4(B) = (\sin^2(B) + \cos^2(B))^2 - 2\sin^2(B)\cos^2(B) \][/tex]
7. Recall the Pythagorean Identity:
We know that:
[tex]\[ \sin^2(B) + \cos^2(B) = 1 \][/tex]
So,
[tex]\[ (\sin^2(B) + \cos^2(B))^2 = 1 \][/tex]
Therefore:
[tex]\[ \sin^4(B) + \cos^4(B) = 1 - 2\sin^2(B)\cos^2(B) \][/tex]
8. Conclusion:
We have shown that:
[tex]\[ \sin^4(B) + \cos^4(B) = 1 - 2\sin^2(B)\cos^2(B) \][/tex]
Which matches the simplified right-hand side:
[tex]\[ 1 - \frac{1}{2} \sin^2(2B) \][/tex]
Thus, the identity:
[tex]\[ \sin^4(B) + \cos^4(B) = 1 - \frac{1}{2} \sin^2(2B) \][/tex]
is verified.
