Bab 3: Trigonometri

$1.150^{\circ }=3\times 360^{\circ }+70^{\circ }=1.080^{\circ }+70^{\circ }$ $\cos 1.150^{\circ }=\cos 70^{\circ }$ (Kuadran I)

$1.900^{\circ }=5\times 360^{\circ }+100^{\circ }=1.800^{\circ }+100^{\circ }$ $\sin 100^{\circ }=\sin (180^{\circ }-80^{\circ })=\sin 80^{\circ }$ (positif, kuadran II)

$1.000^{\circ }=2\times 360^{\circ }+280^{\circ }$ $\tan 280^{\circ }=\tan (360^{\circ }-80^{\circ })=-\tan 80^{\circ }$ (Kuadran IV)

$\tan (-400^{\circ })=-\tan 400^{\circ }$ $-\tan (400^{\circ }-360^{\circ })=-\tan 40^{\circ }$

$\cos (-420^{\circ })=\cos 420^{\circ }$ $\cos (420^{\circ }-360^{\circ })=\cos 60^{\circ }$

$\sin 150^{\circ }=\sin (180^{\circ }-30^{\circ })=\frac{1}{2}$ $\cos 30^{\circ }=\frac{1}{2}\sqrt{3}$ $\frac{1}{2}\times \frac{1}{2}\sqrt{3}=\frac{1}{4}\sqrt{3}$

$\tan 225^{\circ }=\tan (180^{\circ }+45^{\circ })=1$ $\tan 120^{\circ }=\tan (180^{\circ }-60^{\circ })=-\sqrt{3}$ $1-(-\sqrt{3})=1+\sqrt{3}$

$\sin 420^{\circ }=\sin (420^{\circ }-360^{\circ })=\sin 60^{\circ }=\frac{1}{2}\sqrt{3}$ $\cos 480^{\circ }=\cos (480^{\circ }-360^{\circ })=\cos 120^{\circ }=-\frac{1}{2}$ $\frac{1}{2}\sqrt{3}+(-\frac{1}{2})=\frac{1}{2}\sqrt{3}-\frac{1}{2}$

$\sin 120^{\circ }=\frac{1}{2}\sqrt{3}$ $\cos 210^{\circ }=-\frac{1}{2}\sqrt{3}$ $\frac{\frac{1}{2}\sqrt{3}}{-\frac{1}{2}\sqrt{3}}=-1$

$\sin \theta =\frac{p}{1}\to depan=p,miring=1$ $samping=\sqrt{1^2-p^2}=\sqrt{1-p^2}$ Kuadran II, tan negatif: $\tan \theta =-\frac{p}{\sqrt{1-p^2}}$

$\cos \theta =\frac{k}{1}\to samping=k,miring=1$ $depan=\sqrt{1-k^2}$ Kuadran II, sin positif: $\sin \theta =\frac{\sqrt{1-k^2}}{1}=\sqrt{1-k^2}$

$\tan x=-1\to x=135^{\circ }$ (tumpul/kuadran II) $\sin 135^{\circ }=\frac{1}{2}\sqrt{2}$

$\sin 100^{\circ }=\sin (90^{\circ }+10^{\circ })=\cos 10^{\circ }$ $\cos 10^{\circ }=p$

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

$\sin \alpha =\frac{3}{5}\to depan=3,miring=5$ $samping=\sqrt{5^2-3^2}=4$ $\cos \alpha =\frac{4}{5}$ $5\times \frac{4}{5}=4$

$\cos 300^{\circ }=\cos (360^{\circ }-60^{\circ })=\cos 60^{\circ }=\frac{1}{2}$ (Kuadran IV, cos positif)

$\frac{\tan x}{\sec x}=\frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}}=\frac{\sin x}{\cos x}\times \frac{\cos x}{1}=\sin x$

$\tan A=\frac{5}{12}\to \sin A=\frac{5}{13},\cos A=\frac{12}{13}$ $\tan B=\frac{3}{4}\to \sin B=\frac{3}{5},\cos B=\frac{4}{5}$ $\sin (A-B)=(\frac{5}{13}\cdot \frac{4}{5})-(\frac{12}{13}\cdot \frac{3}{5})=\frac{20}{65}-\frac{36}{65}=-\frac{16}{65}$

$\sin A=\frac{3}{5}$ (Kuadran II) $\to \cos A=-\frac{4}{5}$ $2\sin A\cos A=2(\frac{3}{5})(-\frac{4}{5})=-\frac{24}{25}$

$\cos A=\frac{4}{5}\to \sin A=\frac{3}{5}$ $\sin B=\frac{12}{13}\to \cos B=-\frac{5}{13}$ (Kuadran II) $\cos (A+B)=(\frac{4}{5}\cdot -\frac{5}{13})-(\frac{3}{5}\cdot \frac{12}{13})=-\frac{20}{65}-\frac{36}{65}=-\frac{56}{65}$

$\tan A=\frac{7}{24},\tan B=\frac{3}{4}=\frac{18}{24}$ $\tan (A-B)=\frac{\frac{7}{24}-\frac{18}{24}}{1+(\frac{7}{24}\cdot \frac{3}{4})}=\frac{-\frac{11}{24}}{\frac{96+21}{96}}=\frac{-11}{24}\cdot \frac{96}{117}=-\frac{44}{117}$

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

$\sin A=\frac{4}{5},\cos B=\frac{12}{13}$ $\sin (A+B)=(\frac{4}{5}\cdot \frac{12}{13})+(-\frac{3}{5}\cdot \frac{5}{13})=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}$

$\tan x=\frac{1}{a}\to de=1,sa=a,mi=\sqrt{a^2+1}$ $\cos 2x=1-2{\sin }^{2}x=\frac{s{a}^2-d{e}^2}{m{i}^2}=\frac{a^2-1}{a^2+1}$

$\cot x=\frac{1}{k}\to \tan x=k\to de=k,sa=1,mi=\sqrt{k^2+1}$ $2\sin x\cos x=2(\frac{k}{\sqrt{k^2+1}})(\frac{1}{\sqrt{k^2+1}})=\frac{2k}{k^2+1}$