تمارين (2-4)

(1)- أوجد tanx, cosx, sinx إذا علمت أن الضلع النهائي للزاوية (x) الموجهة في الوضع القياسي يقطع دائرة الوحدة في النقط المثلثية الآتية:

أ- $(\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}})$

$\begin{array}{rl}& (\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}) , \sin \mathrm{x}=\frac{-2}{\sqrt{5}} , \cos \mathrm{x}=\frac{1}{\sqrt{5}}\\ & \tan \mathrm{x}=\frac{\sin \mathrm{x}}{\cos \mathrm{x}}=\frac{-2}{\sqrt{5}}÷\frac{1}{\sqrt{5}}=\frac{-2}{\sqrt{5}}\times \sqrt{5}=2\end{array}$

ب- $(-\frac{\sqrt{3}}{3},\frac{\sqrt{6}}{3})$

$\begin{array}{rl}& (\frac{-\sqrt{3}}{2},\frac{\sqrt{6}}{3}) , \sin x=\frac{\sqrt{6}}{3} , \cos x=\frac{-\sqrt{3}}{2}\\ & \tan x=\frac{\sin x}{\cos x}=\frac{\sqrt{6}}{3}÷\frac{-\sqrt{3}}{2}=\frac{\sqrt{6}}{3}\times \frac{-2}{\sqrt{3}}=\frac{-2\sqrt{3}\sqrt{2}}{3\sqrt{3}}=\frac{-2\sqrt{2}}{3\sqrt{3}}\end{array}$

ج- $(\frac{1}{2},\frac{\sqrt{3}}{2})$

$\begin{array}{rl}& (\frac{1}{2},\frac{\sqrt{3}}{2}) , \sin x=\frac{\sqrt{3}}{2} , \cos x=\frac{1}{2}\\ & \tan x=\frac{\sin x}{\cos x}=\frac{\sqrt{3}}{2}÷\frac{1}{2}=\frac{\sqrt{3}}{2}\times 2=\sqrt{3}\end{array}$

د- $(-0.6,-0.8)$

$\begin{array}{rl}& (0.88,-0.48) , \sin x=-0.48 , \cos x=0.88\\ & \tan x=\frac{\sin x}{\cos x}=\frac{-0.48}{0.88}=-0.54\end{array}$

(2)- جد ما يأتي:

أ- $\sin \left(30\mathrm{\Pi }\right)$

$\sin \left(30\mathrm{\pi }\right)=\sin (30\mathrm{\pi }-30\mathrm{\pi })=\sin 0=0$

ب- $\cos (-13\mathrm{\Pi }/6)$

$\begin{array}{rl}& \cos (-\frac{13\mathrm{\pi }}{6})=\cos (\frac{24\mathrm{\pi }}{6}-\frac{13\mathrm{\pi }}{6})=\cos \frac{11\mathrm{\pi }}{6}(30\times 11={330}^{\circ })\\ & \cos \frac{11\mathrm{\pi }}{6}=\cos (2\mathrm{\pi }-\frac{\mathrm{\pi }}{6})=\cos \frac{\mathrm{\pi }}{6}=\frac{\sqrt{3}}{2}\end{array}$

ج- $\tan (4\mathrm{\Pi }/3)$

$\tan \left(\frac{4\mathrm{\pi }}{3}\right)=\tan \frac{4\mathrm{\pi }}{3}=\tan (\mathrm{\pi }+\frac{\mathrm{\pi }}{3})=\sqrt{3}$

د- $\cos \left(30\mathrm{\Pi }\right)$

$\cos (30\mathrm{\pi }-30\mathrm{\pi })=\cos 0(1,0)=1$

(3)- جد قيمة ما يأتي:

أ- ${\sin }^{2}3+{\cos }^{2}3$

${\sin }^{2}3+{\cos }^{2}3=1$

ب- ${\cos }^2\mathrm{\Pi }/6-{\sin }^2\mathrm{\Pi }/6$

$\begin{array}{rl}{\cos }^2\frac{\mathrm{\pi }}{6}-{\sin }^2\frac{\mathrm{\pi }}{6}& ={\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2\\ & =\frac{3}{4}-\frac{1}{4}=\frac{1}{2}\end{array}$

(4)- تحقق مما يأتي:

$\sin \frac{\mathrm{\Pi }}{6}\cos \frac{\mathrm{\Pi }}{3}+\cos \frac{\mathrm{\Pi }}{6}\sin \frac{\mathrm{\Pi }}{3}=\sin \frac{\mathrm{\Pi }}{2}$

$\begin{array}{rl}& \sin \frac{\mathrm{\pi }}{6}\cos \frac{\mathrm{\pi }}{3}+\cos \frac{\mathrm{\pi }}{6}\sin \frac{\mathrm{\pi }}{3}=\sin \frac{\mathrm{\pi }}{2}\\ & \frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}=1\\ & \sin \frac{\mathrm{\pi }}{2}=1\end{array}$

الطرف الأيمن يساوي الطرف الأيسر.