تمارين (4-4)

(1)- إذا كان $\mathrm{\theta } , \sin \mathrm{\theta }=-8/17$ تقع في الربع الثالث فجد:

$\cos \mathrm{\theta } , \cos (3\mathrm{\Pi }/2-\mathrm{\theta }) , \sin (\mathrm{\Pi }/2+\mathrm{\theta })$

$\begin{array}{rl}& \cos \mathrm{\theta }=\frac{15}{17}\\ & \cos \mathrm{\theta }=\frac{-15}{17}\\ & \cos (\frac{3\mathrm{\pi }}{2}-\mathrm{\theta })=\cos (270-\mathrm{\theta })=-\sin \mathrm{\theta }=-\left(\frac{-8}{17}\right)=\frac{8}{17}\\ & \sin (\frac{\mathrm{\pi }}{2}+\mathrm{\theta })=\sin (90+\mathrm{\theta })=\cos \mathrm{\theta }=\frac{15}{17}\\ & =\frac{-15}{17}\end{array}$

(2)- إذا كان ${270}^{\circ }<\mathrm{\beta }<{360}^{\circ } , \cos \mathrm{\beta }=0.8$ فجد:

$\sin \mathrm{\beta } , \cos ({270}^{\circ }+\mathrm{\beta }) , \cos ({270}^{\circ }-\mathrm{\beta })$

$\begin{array}{rl}& {\sin }^2\mathrm{B}=1-\cos {\mathrm{B}}^2=1-{\left(\frac{8}{10}\right)}^2=1-\frac{64}{100}=\frac{100-64}{100}=\frac{36}{100}\\ & \sin \mathrm{B}=\pm \frac{6}{10}\\ & \sin \mathrm{B}=\frac{-6}{10}\end{array}$

$\begin{array}{rl}& \cos (270+\mathrm{B})=\sin \mathrm{B}=\frac{-6}{10}=\frac{-3}{5}\\ & \cos (270-\mathrm{B})=-\sin \mathrm{B}=-\left(\frac{-6}{10}\right)=\frac{6}{10}=\frac{3}{5}\end{array}$

(3)- إذا كان ${90}^{\circ }<\mathrm{\alpha }<{180}^{\circ } , \sin \mathrm{\alpha }=24/25$ فاحسب قيمة:

$\sin ({90}^{\circ }-\mathrm{\alpha })-\cos ({180}^{\circ }-\mathrm{\alpha })+\cos {120}^{\circ }$

$\begin{array}{rl}& \sin (90-\propto )-\cos (180-\propto )+\cos {120}^{\circ }=\mathrm{x}\\ & \mathrm{x}=\cos \propto -(-\cos \propto )+\cos (180-60)\\ & \mathrm{x}=\cos \propto +\cos \propto -\cos 60\\ & \mathrm{x}=2\cos \propto -\frac{1}{2}\end{array}$

$\begin{array}{rl}& {\cos }^2\propto =1-{\sin }^2\propto =1-{\left(\frac{24}{25}\right)}^2=1-\frac{576}{625}=\frac{625-576}{625}\\ & =\frac{49}{625}\Rightarrow \cos \propto =\frac{-7}{25}\end{array}$

$\mathrm{x}=2\left(\frac{-7}{25}\right)-\frac{1}{2}\Rightarrow \mathrm{x}=-\frac{14}{25}-\frac{1}{2}=\frac{-28-25}{50}=\frac{-53}{50}$

(4)- أثبت أنه $\cos (\mathrm{\Pi }/2+\mathrm{\theta })\cos (\mathrm{\Pi }/2-\mathrm{\theta })-\sin (\mathrm{\Pi }+\mathrm{\theta })\sin (\mathrm{\Pi }-\mathrm{\theta })=0$

$\begin{array}{rl}& \cos (\frac{\mathrm{\pi }}{2}+\mathrm{\theta })\cos (\frac{\mathrm{\pi }}{2}-\mathrm{\theta })-\sin (\mathrm{\pi }+\mathrm{\theta })\sin (\mathrm{\pi }-\mathrm{\theta })=0\\ & \sin \mathrm{\theta sin}\mathrm{\theta }-\sin \mathrm{\theta sin}\mathrm{\theta }={\sin }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }=0\end{array}$

الطرف الأيمن يساوي الطرف الأيسر.

(5)- حدد الربع الذي تقع فيه الزاوية $\mathrm{\alpha }$ إذا كان:

أ- $\sin \propto >0 , \cos \propto >0$

في الربع الأول.

ب- $\sin \propto >0 , \cos \propto <0$

في الربع الثاني.

ج- $\sin \propto <0 , \cos \propto <0$

في الربع الثالث.

د- $\sin \propto <0 , \cos \propto >0$

في الربع الرابع.

(6)- أي العبارات الآتية صحيحة وأيها خاطئة؟

1. $\sin {270}^{\circ }=2\sin {30}^{\circ }$ (خطأ)
2. $\sin {90}^{\circ }=2\cos {60}^{\circ }$ (صح)
3. $\cos {150}^{\circ }=1/2\tan {120}^{\circ }$ (صح)
4. $\cos ({30}^{\circ }+{60}^{\circ })=\cos {30}^{\circ }+\cos {60}^{\circ }$ (خطأ)

(7)- أثبت أن:

أ- $\sin ({90}^{\circ }+\propto )+\cot ({270}^{\circ }-\propto )+\cos ({180}^{\circ }+\propto )=\tan \propto$

$\begin{array}{r}\cos \propto +\tan \propto -\cos \mathrm{\alpha }=\tan \mathrm{\alpha }\\ \cos \mathrm{\alpha }-\cos \mathrm{\alpha }+\tan \mathrm{\alpha }=\tan \propto \\ 0+\tan \mathrm{\alpha }=\tan \propto \\ \tan \propto =\tan \propto \end{array}$

ب- ${\sin }^2{135}^{\circ }=1/2(1-\cos {270}^{\circ })$

$\begin{array}{rl}{\sin }^2(180-45)& =\frac{1}{2}(1-0)\\ {\sin }^2{45}^{\circ }& =\frac{1}{2}\\ {\left(\frac{1}{\sqrt{2}}\right)}^2& =\frac{1}{2}\\ \frac{1}{2}& =\frac{1}{2}\end{array}$