Evaluation of Trigonometric Function at a Specific Angle
<p>To solve sin(105°) + cos(105°), we can use the sum-to-product identities.</p>
<p>The sum-to-product identities tell us that:</p>
<p>\sin(\alpha) + \sin(\beta) = 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right)</p>
<p>and</p>
<p>\cos(\alpha) + \cos(\beta) = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right)</p>
<p>However, we need to express sin(105°) as a sine term and cos(105°) as a cosine term to use these identities directly. We can use the co-function identity \sin(\theta) = \cos(90° - \theta) to express \sin(105°) as a cosine:</p>
<p>\sin(105°) = \cos(90° - 105°) = \cos(-15°)</p>
<p>The trigonometric function cosine is even, so:</p>
<p>\cos(-15°) = \cos(15°)</p>
<p>Now we have the expression with both sine and cosine:</p>
<p>\sin(105°) + \cos(105°) = \cos(15°) + \cos(105°)</p>
<p>Using the sum-to-product identities with \alpha = 15° and \beta = 105°, we can combine these terms:</p>
<p>2 \cos\left(\frac{15° + 105°}{2}\right) \cos\left(\frac{15° - 105°}{2}\right) = 2 \cos(60°) \cos(-45°)</p>
<p>\cos(60°) = \frac{1}{2}</p>
<p>\cos(-45°) = \cos(45°) = \frac{\sqrt{2}}{2}</p>
<p>Therefore:</p>
<p>2 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \sqrt{2}</p>
<p>The solution is:</p>
<p>\sin(105°) + \cos(105°) = \sqrt{2}</p>
