Pregunta: If sinθ=94,0<θ<2π, find the exact value of each of the following. (a) sin(2θ) (b) cos(2θ) (c) sin2θ (d) cos2θ (a) sin(2θ)= (Type an exact answer, using radicals as needed.) (b) cos(2θ)= (Type an exact answer, using radicals as needed.) (c) sin2θ= (Type an exact answer, using radicals as needed.) (d) cos2θ=Find the exact value of the expression.

If $\sin \theta =\frac{4}{9},0<\theta <\frac{\pi }{2}$, find the exact value of each of the following. (a) $\sin (2\theta )$ (b) $\cos (2\theta )$ (c) $\sin \frac{\theta }{2}$ (d) $\cos \frac{\theta }{2}$ (a) $\sin (2\theta )=$ (Type an exact answer, using radicals as needed.) (b) $\cos (2\theta )=$ (Type an exact answer, using radicals as needed.) (c) $\sin \frac{\theta }{2}=$ (Type an exact answer, using radicals as needed.) (d) $\cos \frac{\theta }{2}=$ Find the exact value of the expression. $\tan \left[\frac{\pi }{3}+{\sin }^{-1}\frac{12}{13}\right]$ $\tan \left[\frac{\pi }{3}+{\sin }^{-1}\frac{12}{13}\right]=$ (Type an exact answer. Simplify your answer, including any Establish the identity $\tan (\pi +\theta )=\tan \theta$. Which of the following four statements establishes the identity? A. $\tan (\pi +\theta )=\frac{\tan \pi -\tan \theta }{1-\tan \pi \tan \theta }=\frac{1-\tan \theta }{1-(1)\cdot \tan \theta }=\tan \theta$ B. $\tan (\pi +\theta )=\frac{\tan \pi +\tan \theta }{1+\tan \pi \tan \theta }=\frac{0+\tan \theta }{1+0\cdot \tan \theta }=\tan \theta$ C. $\tan (\pi +\theta )=\frac{\tan \pi +\tan \theta }{1-\tan \pi \tan \theta }=\frac{0+\tan \theta }{1-0\cdot \tan \theta }=\tan \theta$ D. $\tan (\pi +\theta )=\frac{\tan \pi -\tan \theta }{1+\tan \pi \tan \theta }=\frac{1+\tan \theta }{1+1\cdot \tan \theta }=\tan \theta$ Establish the identify. $\cos \left(\frac{3x}{2}+\theta \right)=\sin \theta$ Choose the sequence of steps below that verifies the identity. A. $\cos \left(\frac{3\pi }{2}+\theta \right)=\cos \frac{3\pi }{2}\cos \theta -\sin \frac{3\pi }{2}\sin \theta =(0)\cos \theta -(-1)\sin \theta =\sin \theta$ 8. $\cos \left(\frac{3\pi }{2}+\theta \right)=\sin \frac{3\pi }{2}\cos \theta -\cos \frac{3\pi }{2}\sin \theta =(-1)\cos \theta +(0)\sin \theta =\sin \theta$ c. $\cos \left(\frac{3\pi }{2}+\theta \right)=\sin \frac{3\pi }{2}\cos \theta +\cos \frac{3\pi }{2}\sin \theta =(0)\cos \theta -(0)\sin \theta =\sin \theta$ D. $\cos \left(\frac{3x}{2}+\theta \right)=\cos \frac{3t}{2}\cos \theta +\sin \frac{3\pi }{2}\sin \theta =(0)\cos \theta +(-1)\sin \theta =\sin \theta$ Establish the identity. $\sin (\pi -\theta )=\sin \theta$ Choose the sequence of steps below that verifies the identity. A. $\sin (\pi -\theta )=\cos \pi \cos \theta +\sin \pi \sin \theta =(0)\cos \theta -(0)\sin \theta =\sin \theta$ B. $\sin (\pi -\theta )=\sin \pi \cos \theta -\cos \pi \sin \theta =(0)\cos \theta -(-1)\sin \theta =\sin \theta$ C. $\sin (\pi -\theta )=\cos \pi \cos \theta -\sin \pi \sin \theta =(-1)\cos \theta +(0)\sin \theta =\sin \theta$ D. $\sin (\pi -\theta )=\sin \pi \cos \theta +\cos \pi \sin \theta =(0)\cos \theta +(-1)\sin \theta =\sin \theta$ Establish the identity $\sin \left(\frac{3\pi }{2}-\theta \right)=-\cos \theta$. Which of the following four statements establishes the identity? A. $\sin \left(\frac{3\pi }{2}-\theta \right)=\sin \frac{3\pi }{2}\sin \theta +\cos \frac{3\pi }{2}\cos \theta =(-1)\cos \theta +(0)\sin \theta =-\cos \theta$ B. $\sin \left(\frac{3\pi }{2}-\theta \right)=\sin \frac{3\pi }{2}\cos \theta -\cos \frac{3\pi }{2}\sin \theta =(-1)\cos \theta -(0)\sin \theta =-\cos \theta$ C. $\sin \left(\frac{3\pi }{2}-\theta \right)=\sin \frac{3\pi }{2}\sin \theta -\cos \frac{3\pi }{2}\cos \theta =(0)\sin \theta -(-1)\cos \theta =-\cos \theta$ D. $\sin \left(\frac{3\pi }{2}-\theta \right)=\sin \frac{3\pi }{2}\cos \theta +\cos \frac{3\pi }{2}\sin \theta =(-1)\cos \theta +(0)\sin \theta =-\cos \theta$ Establish the identity. $1+{\tan }^2(-\theta )={\sec }^2\theta$ Which of the following four statements establishes the identity? A. $1+{\tan }^2(-\theta )=1+(-\tan \theta {)}^2=1-{\tan }^2\theta ={\sec }^2\theta$ B. $1+{\tan }^2(-\theta )=1-{\tan }^2(-\theta )=1+{\tan }^2\theta ={\sec }^2\theta$ C. $1+{\tan }^2(-\theta )=1+(-\tan \theta {)}^2=1+{\tan }^2\theta ={\sec }^2\theta$ D. $1+{\tan }^2(-\theta )=(\sec \theta {)}^2-1={\sec }^2\theta -1={\sec }^2\theta$