We are given: \(\sin \theta + \cos \theta = x\)
We need to prove: \(\sin^4 \theta + \cos^4 \theta = \dfrac{2 - (x^2 - 1)^2}{2}\)
Step 1: Use the identity:
\[
\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta
\]
Since \(\sin^2 \theta + \cos^2 \theta = 1\), we get:
\[
\sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \quad \text{(i)}
\]
Now, square the given expression:
\[
(\sin \theta + \cos \theta)^2 = x^2 \Rightarrow \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = x^2
\Rightarrow 1 + 2\sin \theta \cos \theta = x^2
\Rightarrow \sin \theta \cos \theta = \dfrac{x^2 - 1}{2}
\]
Now square both sides:
\[
\sin^2 \theta \cos^2 \theta = \left( \dfrac{x^2 - 1}{2} \right)^2 = \dfrac{(x^2 - 1)^2}{4}
\]
Substitute into (i):
\[
\sin^4 \theta + \cos^4 \theta = 1 - 2 \cdot \dfrac{(x^2 - 1)^2}{4} = 1 - \dfrac{(x^2 - 1)^2}{2}
= \dfrac{2 - (x^2 - 1)^2}{2}
\]
Hence proved.