Question

If \(A=45°,\ B=60°,\) then \(sin\ A+cos\ B\)

• A. \(\frac {2-\sqrt 2}{2\sqrt2}\)
• B. \(\frac {2+\sqrt 2}{2}\)
• C. \(\frac {2+\sqrt 2}{\sqrt2}\)
• D. \(\frac {2+\sqrt 2}{2\sqrt2}\)

Answer 1

The Correct Option is D

To solve the problem, we need to find the value of \( \sin A + \cos B \) given that \( A = 45^\circ \) and \( B = 60^\circ \).

1. Use Standard Trigonometric Values:
We know the standard values:
\( \sin 45^\circ = \frac{1}{\sqrt{2}} \)

\( \cos 60^\circ = \frac{1}{2} \)

2. Substitute the Values:
\[ \sin A + \cos B = \sin 45^\circ + \cos 60^\circ = \frac{1}{\sqrt{2}} + \frac{1}{2} \]

3. Take LCM to Simplify:
To simplify the sum:
\[ \frac{1}{\sqrt{2}} + \frac{1}{2} = \frac{2 + \sqrt{2}}{2\sqrt{2}} \]

Final Answer:
The value of \( \sin A + \cos B \) is \( \frac{2 + \sqrt{2}}{2\sqrt{2}} \).