Given $\alpha \in ( 0 , \pi )$ and $3 \cos 2 \alpha - 8 \cos \alpha = 5$, then $\sin \alpha =$
A. $\frac { \sqrt { 5 } } { 3 }$
B. $\frac { 2 } { 3 }$
C. $\frac { 1 } { 3 }$
D. $\frac { \sqrt { 5 } } { 9 }$

``$\sin^{2} \alpha + \sin^{2} \beta = 1$'' is ``$\cos \alpha + \cos \beta = 0$'' a
A. sufficient but not necessary condition
B. necessary but not sufficient condition
C. necessary and sufficient condition
D. neither sufficient nor necessary condition

If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^ { 2 } A + \sin ^ { 2 } B + \sin ^ { 2 } C = 2 \left( \cos ^ { 2 } A + \cos ^ { 2 } B + \cos ^ { 2 } C \right) ,$$ prove that the triangle must have a right angle.

If
$$\frac { \sin ^ { 4 } x } { 2 } + \frac { \cos ^ { 4 } x } { 3 } = \frac { 1 } { 5 } ,$$
then
(A) $\quad \tan ^ { 2 } x = \frac { 2 } { 3 }$
(B) $\quad \frac { \sin ^ { 8 } x } { 8 } + \frac { \cos ^ { 8 } x } { 27 } = \frac { 1 } { 125 }$
(C) $\quad \tan ^ { 2 } x = \frac { 1 } { 3 }$
(D) $\frac { \sin ^ { 8 } x } { 8 } + \frac { \cos ^ { 8 } x } { 27 } = \frac { 2 } { 125 }$

If $\tan A$ and $\tan B$ are the roots of the quadratic equation, $3 x ^ { 2 } - 10 x - 25 = 0$ then the value of $3 \sin ^ { 2 } ( A + B ) - 10 \sin ( A + B ) \cdot \cos ( A + B ) - 25 \cos ^ { 2 } ( A + B )$ is
(1) 25
(2) - 25
(3) - 10
(4) 10

If $\sin ^ { 4 } \alpha + 4 \cos ^ { 4 } \beta + 2 = 4 \sqrt { 2 } \sin \alpha \cos \beta , \alpha , \beta \in [ 0 , \pi ]$, then $\cos ( \alpha + \beta ) - \cos ( \alpha - \beta )$ is equal to
(1) - 1
(2) $- \sqrt { 2 }$
(3) $\sqrt { 2 }$
(4) 0

If $\sin x = - \frac { 3 } { 5 }$, where $\pi < x < \frac { 3 \pi } { 2 }$, then $80 \left( \tan ^ { 2 } x - \cos x \right)$ is equal to
(1) 108
(2) 109
(3) 18
(4) 19

In Figure 1, a parallelogram with two sides of lengths 4 units and 8 units and the angle between these sides measuring $x$ degrees is given. In Figure 2, a parallelogram with two sides of lengths 4 units and 6 units and the angle between these sides measuring $2x$ degrees is given.
If the area of the parallelogram in Figure 1 is 24 square units, what is the area of the parallelogram in Figure 2 in square units?
A) $6\sqrt{7}$ B) $7\sqrt{7}$ C) $8\sqrt{7}$ D) $9\sqrt{7}$ E) $10\sqrt{7}$