Question

Given that $\sin \theta + \cos \theta = x$, prove that $\sin^4 \theta + \cos^4 \theta = \dfrac{2 - (x^2 - 1)^2}{2}$.

Hint:

Convert powers to squares using algebraic identities, and square sum expressions carefully.

Answer 1

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 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.