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

a) $\frac{\sqrt{3}}{2}$ b) $-\frac{1}{2}$ c) $-\sqrt{3}$

$\cos \theta =-\frac{2\sqrt{5}}{5},\sin \theta =\frac{\sqrt{5}}{5},\tan \theta =-\frac{1}{2},\csc \theta =\sqrt{5},\sec \theta =-\frac{\sqrt{5}}{2},\cot \theta =-2$

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

$\frac{\sqrt{2-\sqrt{2}}}{2}$

$\frac{\sqrt{2-\sqrt{3}}}{2}$

$2+\sqrt{3}$

$-1-\sqrt{2}$

a) $\frac{3\sqrt{13}}{13}$ b) $-\frac{2\sqrt{13}}{13}$ c) $-\frac{3}{2}$

a) $\frac{\sqrt{10}}{4}$ b) $\frac{\sqrt{6}}{4}$ c) $\frac{\sqrt{15}}{3}$

$\frac{120}{169},–\frac{119}{169},–\frac{120}{119}$

$\frac{2\sqrt{13}}{13},\frac{3\sqrt{13}}{13},\frac{2}{3}$

$\cos ({74}^{\circ })$

$\cos (18x)$

$3\sin (10x)$

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

$\begin{array}{l}\frac{\sin \left(2\theta \right)}{1+\cos \left(2\theta \right)}{\tan }^2\theta =\frac{2\sin \left(\theta \right)\cos \left(\theta \right)}{1+{\cos }^2\theta -{\sin }^2\theta }{\tan }^2\theta =\\ \frac{2\sin \left(\theta \right)\cos \left(\theta \right)}{2{\cos }^2\theta }{\tan }^2\theta =\frac{\sin \left(\theta \right)}{\cos \theta }{\tan }^2\theta =\\ \cot \left(\theta \right){\tan }^2\theta =\tan \theta \end{array}$

$\frac{1+\cos (12x)}{2}$

$\frac{3+\cos (12x)-4\cos (6x)}{8}$

$\frac{2+\cos (2x)-2\cos (4x)-\cos (6x)}{32}$

$\frac{3+\cos (4x)-4\cos (2x)}{3+\cos (4x)+4\cos (2x)}$

$\frac{1-\cos (4x)}{8}$

$\frac{3+\cos (4x)-4\cos (2x)}{4(\cos (2x)+1)}$

$\frac{\left(1+\cos \left(4x\right)\right)\sin x}{2}$

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

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

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

$\begin{array}{l}\sin (x+2x)=\sin x\cos (2x)+\sin (2x)\cos x\hfill \\ =\sin x({\cos }^{2}x-{\sin }^{2}x)+2\sin x\cos x\cos x\hfill \\ =\sin x{\cos }^{2}x-{\sin }^{3}x+2\sin x{\cos }^{2}x\hfill \\ =3\sin x{\cos }^{2}x-{\sin }^{3}x\hfill \end{array}$

$\begin{array}{l}\frac{1+\cos (2t)}{\sin (2t)-\cos t}=\frac{1+2{\cos }^{2}t-1}{2\sin t\cos t-\cos t}\hfill \\ =\frac{2{\cos }^{2}t}{\cos t(2\sin t-1)}\hfill \\ =\frac{2\cos t}{2\sin t-1}\hfill \end{array}$

$\begin{array}{l}\left({\cos }^2(4x)-{\sin }^2(4x)-\sin (8x))({\cos }^2(4x)-{\sin }^2(4x)+\sin (8x)\right)=\hfill \\ =\left(\cos (8x)-\sin (8x))(\cos (8x)+\sin (8x)\right)\hfill \\ ={\cos }^2(8x)-{\sin }^2(8x)\hfill \\ =\cos (16x)\hfill \\ \hfill \end{array}$