Question

If \( \sin x + \cos x = \sqrt{2} \), then the value of \( \sin x \cos x \) is equal to:

• A. $1$
• B. $\frac{1}{2}$
• C. $2$
• D. $\sqrt{2}$
• E. $\frac{1}{2\sqrt{2}}$

Hint:

The expression $\sin x + \cos x$ can also be written as $\sqrt{2} \sin(x + 45^\circ)$. If this equals $\sqrt{2}$, then $\sin(x + 45^\circ) = 1$, which means $x = 45^\circ$. Substituting $x=45^\circ$ into $\sin x \cos x$ gives $(1/\sqrt{2}) \times (1/\sqrt{2}) = 1/2$.

Answer 1

The Correct Option is B

Concept: To find the product $\sin x \cos x$ from the sum $\sin x + \cos x$, we utilize the algebraic identity $(a+b)^2 = a^2 + b^2 + 2ab$. In trigonometry, this is particularly useful because of the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, which allows us to simplify the squared sum into a constant plus the product term we are seeking.

Step 1: Square both sides of the given equation to eliminate the radical.
Given the equation: $\sin x + \cos x = \sqrt{2}$. By squaring both sides of the equality, we maintain the balance of the equation: \[ (\sin x + \cos x)^2 = (\sqrt{2})^2 \] Expanding the left-hand side using the perfect square formula $(A+B)^2 = A^2 + 2AB + B^2$: \[ \sin^2 x + \cos^2 x + 2 \sin x \cos x = 2 \]

Step 2: Apply the fundamental trigonometric identity.
Recall that for any angle $x$, the identity $\sin^2 x + \cos^2 x = 1$ always holds true. We substitute this value into our expanded equation: \[ 1 + 2 \sin x \cos x = 2 \]

Step 3: Isolate the product term $\sin x \cos x$.
Subtract $1$ from both sides of the equation to simplify: \[ 2 \sin x \cos x = 2 - 1 \] \[ 2 \sin x \cos x = 1 \] Finally, divide both sides by $2$ to solve for the product: \[ \sin x \cos x = \frac{1}{2} \]