Question - Calculation of Trigonometric Expression and Centroid Coordinates

\text{For the trigonometric expression:}

\sin(105^\circ) + \cos(105^\circ) = \sin(60^\circ+45^\circ) + \cos(60^\circ+45^\circ)\

\text{Using the sine and cosine addition formulas:}

\sin(A+B) = \sin{A}\cos{B} + \cos{A}\sin{B}

\cos(A+B) = \cos{A}\cos{B} - \sin{A}\sin{B}

\text{Substitute } A = 60^\circ \text{ and } B = 45^\circ:

\sin(60^\circ)\cos(45^\circ) + \cos(60^\circ)\sin(45^\circ) + \cos(60^\circ)\cos(45^\circ) - \sin(60^\circ)\sin(45^\circ)\

\text{Using known values of sine and cosine for } 60^\circ \text{ and } 45^\circ:

\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} + \frac{1}{2}\cdot\frac{\sqrt{2}}{2} + \frac{1}{2}\cdot\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}

\text{Simplify the expression:}

\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}

\text{The terms with } \sqrt{6} \text{ cancel out, so we are left with:}

\frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}

\text{For the centroid coordinates:}

\text{Let the coordinates of } C \text{ be } (x_C, y_C). \text{ The centroid coordinates (G) can be found by:}

G_x = \frac{x_A + x_B + x_C}{3}, G_y = \frac{y_A + y_B + y_C}{3}

\text{Substitute the given values with } G (1,1,1):

1 = \frac{3 + (-1) + x_C}{3}, 1 = \frac{-5 + (-7) + y_C}{3}

\text{Solve for } x_C \text{ and } y_C:

x_C = 3 - (3 + (-1)) = 3 - 2 = 1

y_C = 3 - (-5 -7) = 3 + 12 = 15

\text{So, the coordinates of point C are } (1, 15).

\text{For the trigonometric identity:}

\cot{(\theta)} + \tan{(\theta)} = 2\csc{(\theta)}

\text{Using the identities } \cot{(\theta)} = \frac{1}{\tan{(\theta)}} \text{ and } \csc{(\theta)} = \frac{1}{\sin{(\theta)}}, \text{ we get:}

\frac{1}{\tan{(\theta)}} + \tan{(\theta)} = \frac{2}{\sin{(\theta)}}

\text{Multiply by } \tan{(\theta)}\sin{(\theta)} \text{ to clear the denominators:}

\sin{(\theta)} + \tan^2{(\theta)}\sin{(\theta)} = 2\tan{(\theta)}

\text{Use the identity } \sin^2{(\theta)} + \cos^2{(\theta)} = 1 \text{ to express } \tan{(\theta)} \text{ in terms of } \sin{(\theta)}:

\tan{(\theta)} = \frac{\sin{(\theta)}}{\cos{(\theta)}}, \text{ so }

\sin{(\theta)} + \frac{\sin^3{(\theta)}}{\cos^2{(\theta)}} = 2\frac{\sin{(\theta)}}{\cos{(\theta)}}

\text{Multiply by } \cos^2{(\theta)} \text{ to clear the denominators:}

\sin{(\theta)}\cos^2{(\theta)} + \sin^3{(\theta)} = 2\sin{(\theta)}\cos{(\theta)}

\text{Rearrange and factor out } \sin{(\theta)}:

\sin{(\theta)}(\cos^2{(\theta)} - 2\cos{(\theta)} + \sin^2{(\theta)}) = 0

\text{Since } \sin{(\theta)} = 0 \text{ leads to no solution that satisfies the original equation, focus on:}

\cos^2{(\theta)} - 2\cos{(\theta)} + \sin^2{(\theta)} = 0

\text{Using } \sin^2{(\theta)} = 1 - \cos^2{(\theta)}:

\cos^2{(\theta)} - 2\cos{(\theta)} + (1 - \cos^2{(\theta)}) = 0

-2\cos{(\theta)} + 1 = 0

\cos{(\theta)} = \frac{1}{2}

\text{Hence, the general values of } \theta \text{ can be } 60^\circ + n360^\circ \text{ or } 300^\circ + n360^\circ, \text{ where } n \text{ is an integer.}