Question

If \(\theta = 45^\circ\) then the value of \(\sin \theta + \cos \theta\) is

• A. \(\frac{1}{\sqrt{2}}\)
• B. \(\sqrt{2}\)
• C. \(\frac{1}{2}\)
• D. 1

Hint:

Remember that \(2/\sqrt{2} = \sqrt{2}\). This is a common simplification that can save you a step in calculations.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to evaluate a trigonometric expression by substituting the given angle and using standard trigonometric values.

Step 2: Key Formula or Approach:
We will use the standard values for \(\sin 45^\circ\) and \(\cos 45^\circ\).
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \] \[ \cos 45^\circ = \frac{1}{\sqrt{2}} \]

Step 3: Detailed Explanation:
Substitute \(\theta = 45^\circ\) into the expression \(\sin \theta + \cos \theta\):
\[ \sin 45^\circ + \cos 45^\circ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \] \[ = \frac{2}{\sqrt{2}} \] To simplify, we can rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{2}\):
\[ = \frac{2 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{2\sqrt{2}}{2} \] \[ = \sqrt{2} \] Alternatively, we can write \(2\) as \((\sqrt{2})^2\): \[ \frac{(\sqrt{2})^2}{\sqrt{2}} = \sqrt{2} \]

Step 4: Final Answer:
The value of \(\sin 45^\circ + \cos 45^\circ\) is \(\sqrt{2}\).