Question

If $A = \sin 45^{\circ} + \cos 45^{\circ}$ and $B = \sin 44^{\circ} + \cos 44^{\circ}$ then which of the following is TRUE}

• A. $A > B$
• B. $A < B$
• C. $A = B$
• D. $AB = 1$

Hint:

$\sin x + \cos x$ reaches its maximum value at $x = 45^{\circ}$. Any deviation from $45^{\circ}$ results in a smaller value.

Answer 1

The Correct Option is A

Step 1: Concept
The expression $\sin x + \cos x$ can be rewritten as $\sqrt{2} \sin(x + 45^{\circ})$ using the identity $a \sin x + b \cos x = \sqrt{a^{2}+b^{2}} \sin(x + \alpha)$.

Step 2: Meaning
Let $f(x) = \sin x + \cos x = \sqrt{2} \sin(x + 45^{\circ})$. The function $\sin \theta$ is increasing for $0 < \theta < 90^{\circ}$.

Step 3: Analysis
$A = f(45^{\circ}) = \sqrt{2} \sin(90^{\circ}) = \sqrt{2}$. $B = f(44^{\circ}) = \sqrt{2} \sin(89^{\circ})$. Since $\sin 90^{\circ} > \sin 89^{\circ}$, it follows that $A > B$.

Step 4: Conclusion
By comparing the values using the transformed sine function, $A$ is greater than $B$. Final Answer: (A)