9.3 Double-angle, half-angle, and reduction formulas  (Page 5/8)

If $\tan x=\mathrm{-8},$ and $x$ is in quadrant IV.

$\cos (2\theta )=\frac{3}{5}$ and $\mathrm{90°}\le \theta \le \mathrm{180°}$

$\cos (2\theta )=\frac{1}{\sqrt{2}}$ and $\mathrm{180°}\le \theta \le \mathrm{270°}$

$2\sin \left(\frac{\pi }{4}\right)2\cos \left(\frac{\pi }{4}\right)$

$4\sin \left(\frac{\pi }{8}\right)\cos \left(\frac{\pi }{8}\right)$

$\sin \left(\frac{\pi }{8}\right)$

$\cos \left(-\frac{11\pi }{12}\right)$

$\sin \left(\frac{11\pi }{12}\right)$

$\cos \left(\frac{7\pi }{8}\right)$

$\tan \left(\frac{5\pi }{12}\right)$

$\tan \left(-\frac{3\pi }{12}\right)$

$\tan \left(-\frac{3\pi }{8}\right)$

If $\tan x=-\frac{4}{3},$ and $x$ is in quadrant IV.

If $\sin x=-\frac{12}{13},$ and $x$ is in quadrant III.

If $\csc x=7,$ and $x$ is in quadrant II.

If $\sec x=-4,$ and $x$ is in quadrant II.

Find $\sin \left(2\theta \right),\cos (2\theta ),$ and $\tan (2\theta ).$

Find $\sin (2\alpha ),\cos (2\alpha ),$ and $\tan (2\alpha ).$

Find $\sin \left(\frac{\theta }{2}\right),\cos \left(\frac{\theta }{2}\right),$ and $\tan \left(\frac{\theta }{2}\right).$

Find $\sin \left(\frac{\alpha }{2}\right),\cos \left(\frac{\alpha }{2}\right),$ and $\tan \left(\frac{\alpha }{2}\right).$

${\cos }^2(\mathrm{28°})-{\sin }^2(\mathrm{28°})$

$2{\cos }^2(\mathrm{37°})-1$

$1-2{\sin }^2(\mathrm{17°})$

${\cos }^2(9x)-{\sin }^2(9x)$

$4\sin (8x)\cos (8x)$

$6\sin (5x)\cos (5x)$

${\left(\sin t-\cos t\right)}^2=1-\sin \left(2t\right)$

$\sin \left(2x\right)=-2\sin \left(-x\right)\cos \left(-x\right)$

$\cot x-\tan x=2\cot \left(2x\right)$

$\frac{\sin \left(2\theta \right)}{1+\cos \left(2\theta \right)}{\tan }^2\theta =\tan \theta$

${\cos }^2(5x)$

${\cos }^2(6x)$

${\sin }^4(8x)$

${\sin }^4(3x)$

${\cos }^{2}x{\sin }^{4}x$

${\cos }^{4}x{\sin }^{2}x$

${\tan }^{2}x{\sin }^{2}x$

${\tan }^{4}x$

${\sin }^2(2x)$

${\sin }^{2}x{\cos }^{2}x$

${\tan }^{2}x\sin x$

${\tan }^{4}x{\cos }^{2}x$

${\cos }^{2}x\sin \left(2x\right)$

${\cos }^2\left(2x\right)\sin x$

${\tan }^2\left(\frac{x}{2}\right)\sin x$

$\sin (4x)$

$\cos (4x)$

$\sin \left(2x\right)=\frac{2\tan x}{1+{\tan }^{2}x}$

$\cos (2\alpha )=\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$

$\tan (2x)=\frac{2\sin x\cos x}{2{\cos }^{2}x-1}$

${\left({\sin }^{2}x-1\right)}^2=\cos \left(2x\right)+{\sin }^{4}x$

$\sin \left(3x\right)=3\sin x{\cos }^{2}x-{\sin }^{3}x$

$\cos \left(3x\right)={\cos }^{3}x-3{\sin }^{2}x\cos x$

$\frac{1+\cos \left(2t\right)}{\sin \left(2t\right)-\cos t}=\frac{2\cos t}{2\sin t-1}$

$\sin \left(16x\right)=16\sin x\cos x\cos \left(2x\right)\cos \left(4x\right)\cos \left(8x\right)$

$\cos \left(16x\right)=\left({\cos }^2\left(4x\right)-{\sin }^2\left(4x\right)-\sin \left(8x\right)\right)\left({\cos }^2\left(4x\right)-{\sin }^2\left(4x\right)+\sin \left(8x\right)\right)$