Use the information given about the angle, [tex]$0 \leq \theta\ \textless \ 2 \pi$[/tex], to find the exact value of each trigonometric function. [tex]$\tan \theta=-12, \sin \theta\ \textless \ 0$[/tex]
(a) [tex]$\sin (2 \theta)$[/tex]
(b) [tex]$\cos (2 \theta)$[/tex]
(c) [tex]$\sin \frac{\theta}{2}$[/tex]
(d) [tex]$\cos \frac{\theta}{2}$[/tex]
(e) [tex]$\tan 2 \theta$[/tex]
(f) [tex]$\operatorname { tan } \frac{\theta}{2}$[/tex]
(a) [tex]$\operatorname { s i n } (2 \theta)=$[/tex] $\square$
(Type an exact answer, using radicals as needed.)

Determine the quadrant of θ based on the signs of tan ( θ ) and sin ( θ ) .
Find cos ( θ ) and sin ( θ ) using the given tan ( θ ) and the quadrant.
Use the double angle formulas to find sin ( 2 θ ) , cos ( 2 θ ) , and tan ( 2 θ ) .
Use the half-angle formulas to find sin ( 2 θ ​ ) , cos ( 2 θ ​ ) , and tan ( 2 θ ​ ) .
The final answers are: sin ( 2 θ ) = 145 − 24 ​ , cos ( 2 θ ) = 145 − 143 ​ , sin ( 2 θ ​ ) = 290 145 − 145 ​ ​ ​ , cos ( 2 θ ​ ) = − 290 145 + 145 ​ ​ ​ , tan ( 2 θ ) = 143 24 ​ , and tan ( 2 θ ​ ) = − 12 145 ​ − 1 ​ .

Explanation

Understanding the Problem We are given that tan ( θ ) = − 12 and sin ( θ ) < 0 , with 0 ≤ θ < 2 π . Our goal is to find the exact values of sin ( 2 θ ) , cos ( 2 θ ) , sin ( 2 θ ​ ) , cos ( 2 θ ​ ) , tan ( 2 θ ) , and tan ( 2 θ ​ ) .

Finding Sine and Cosine Since tan ( θ ) < 0 and sin ( θ ) < 0 , θ must be in quadrant IV. In quadrant IV, 0"> cos ( θ ) > 0 . We can think of a right triangle where the opposite side is 12 and the adjacent side is 1. The hypotenuse is 1 2 + 1 2 2 ​ = 145 ​ . Therefore, cos ( θ ) = 145 ​ 1 ​ and sin ( θ ) = 145 ​ − 12 ​ .

Double Angle Formulas Now we use the double angle formulas: sin ( 2 θ ) = 2 sin ( θ ) cos ( θ ) = 2 ⋅ 145 ​ − 12 ​ ⋅ 145 ​ 1 ​ = 145 − 24 ​ cos ( 2 θ ) = cos 2 ( θ ) − sin 2 ( θ ) = ( 145 ​ 1 ​ ) 2 − ( 145 ​ − 12 ​ ) 2 = 145 1 ​ − 145 144 ​ = 145 − 143 ​ tan ( 2 θ ) = 1 − tan 2 ( θ ) 2 tan ( θ ) ​ = 1 − ( − 12 ) 2 2 ( − 12 ) ​ = 1 − 144 − 24 ​ = − 143 − 24 ​ = 143 24 ​

Half Angle Formulas Next, we use the half-angle formulas. Since θ is in quadrant IV, 2 3 π ​ < θ < 2 π , so 4 3 π ​ < 2 θ ​ < π , which means 2 θ ​ is in quadrant II. In quadrant II, 0"> sin ( 2 θ ​ ) > 0 and cos ( 2 θ ​ ) < 0 .
sin ( 2 θ ​ ) = 2 1 − cos ( θ ) ​ ​ = 2 1 − 145 ​ 1 ​ ​ ​ = 2 145 ​ 145 ​ − 1 ​ ​ = 2 145 ​ 145 ​ − 1 ​ ⋅ 145 ​ 145 ​ ​ ​ = 290 145 − 145 ​ ​ ​ cos ( 2 θ ​ ) = − 2 1 + cos ( θ ) ​ ​ = − 2 1 + 145 ​ 1 ​ ​ ​ = − 2 145 ​ 145 ​ + 1 ​ ​ = − 2 145 ​ 145 ​ + 1 ​ ⋅ 145 ​ 145 ​ ​ ​ = − 290 145 + 145 ​ ​ ​ tan ( 2 θ ​ ) = cos ( 2 θ ​ ) sin ( 2 θ ​ ) ​ = − 290 145 + 145 ​ ​ ​ 290 145 − 145 ​ ​ ​ ​ = − 145 + 145 ​ 145 − 145 ​ ​ ​ = − 14 5 2 − 145 ( 145 − 145 ​ ) 2 ​ ​ = − 145 ( 145 − 1 ) ( 145 − 145 ​ ) 2 ​ ​ = − 145 ⋅ 144 ​ 145 − 145 ​ ​ = − 12 145 ​ 145 − 145 ​ ​ = − 12 145 ​ − 1 ​

Final Answer Therefore, we have: sin ( 2 θ ) = 145 − 24 ​ cos ( 2 θ ) = 145 − 143 ​ sin ( 2 θ ​ ) = 290 145 − 145 ​ ​ ​ cos ( 2 θ ​ ) = − 290 145 + 145 ​ ​ ​ tan ( 2 θ ) = 143 24 ​ tan ( 2 θ ​ ) = − 12 145 ​ − 1 ​


Examples
Trigonometric functions are essential in various fields such as physics, engineering, and navigation. For instance, when analyzing the motion of a pendulum or designing electrical circuits, understanding the relationships between angles and their trigonometric values is crucial. In navigation, these functions help determine the position and direction of a ship or aircraft. Moreover, in computer graphics, trigonometric functions are used to create realistic images and animations by manipulating angles and shapes.

Answered by GinnyAnswer | 2025-07-08