إذا كان $\frac{3\mathrm{\Pi }}{2}<x<2\mathrm{\Pi }$ وكانت $\cos \mathrm{x}=\frac{3}{5}$ فأوجد قيمة كل من: $\cos 2\mathrm{x} , \sin 2\mathrm{x} , \tan 2\mathrm{x}$

$$\begin{gathered} {\sin }^2\mathrm{x}=1-{\cos }^2\mathrm{x}=1-\frac{9}{25}=\frac{25}{25}-\frac{9}{25}=\frac{16}{25}\to \sin \mathrm{x}=-\frac{4}{5} \\ \cos 2\mathrm{x}=2{\cos }^2\mathrm{x}-1=2\times \frac{9}{25}-1=\frac{18-25}{25}=-\frac{7}{25} \end{gathered}$$

$\begin{array}{rl}& \sin 2\mathrm{x}=2\sin \mathrm{xcos}\mathrm{x}=2\times \frac{-4}{5}\times \frac{3}{5}=\frac{-24}{25}\\ & \tan \mathrm{x}=\frac{\sin \mathrm{x}}{\cos \mathrm{x}}=-\frac{4}{5}÷\frac{3}{5}=-\frac{4}{5}\times \frac{5}{3}=\frac{-4}{3}\\ & \tan 2\mathrm{x}=\frac{2\tan \mathrm{x}}{1-{\tan }^2\mathrm{x}}=\frac{2\times \frac{-4}{3}}{1-{(-\frac{4}{3})}^2}=\frac{-\frac{8}{3}}{1-\frac{16}{9}}=\frac{-\frac{3}{7}}{-\frac{7}{9}}=\frac{8}{3}÷\frac{7}{9}=\frac{8}{3}\times \frac{9}{7}=\frac{24}{7}\end{array}$

تمارين (5-4)

(1)- ارسم بيان كل من الدوال الآتية، ومن الرسم استنتج كلاً من دورة الدالة وترددها وسعتها:

1) $\mathrm{y}=\sin 3\mathrm{x}\text{on}[0,4\mathrm{\Pi }/3]$

$$\begin{gathered} 2\frac{\mathrm{\pi }}{3} \\ \mathrm{F}=\frac{1}{\frac{2\mathrm{\pi }}{3}}=\frac{3}{2\mathrm{\pi }} \\ \mathrm{A}=1|=|-1\mid =1 \end{gathered}$$

2) $y=-\sin x\text{on}[0,2\mathrm{\Pi }]$

$\mathrm{p}=2\mathrm{\pi } , \mathrm{F}=\frac{1}{2\mathrm{\pi }} , \mathrm{A}=|-1|=\left|1\right|=1$

3) $y=3\sin 2x\text{on}[0,2\mathrm{\Pi }]$

$\begin{array}{rl}& \mathrm{A}=\left|3\right|=|-3|=3\\ & \mathrm{p}=\frac{2\pi }{2}=\pi \\ & \mathrm{F}=\frac{1}{\mathrm{p}}=\frac{1}{\pi }\end{array}$

4) $y=cos2x\text{on}[-\mathrm{\Pi },2\mathrm{\Pi }]$

$$\begin{gathered} \mathrm{p}=\frac{2\mathrm{\pi }}{\mathrm{B}}=\frac{2\mathrm{\pi }}{2}=\mathrm{\pi } \\ \mathrm{F}=\frac{1}{\mathrm{p}}=\frac{1}{\mathrm{\pi }} \\ \mathrm{A}=1 \end{gathered}$$

5) $\mathrm{y}=-2\cos \mathrm{x}\text{on}[-2\mathrm{\Pi },2\mathrm{\Pi }]$

6) $\mathrm{y}=2\cos 3\mathrm{x}\text{on}[0,3\mathrm{\Pi }]$

7) $y=2tanx\text{on}[-\mathrm{\Pi }/2,3\mathrm{\Pi }/2]$

السعة غير محددة.

$\begin{array}{rl}& p=\pi \\ & F=\frac{1}{p}=\frac{1}{\pi }\end{array}$

8) $\mathrm{y}=\tan 2\mathrm{x}\text{on}[0,\mathrm{\Pi }]$

السعة غير محددة.

$\begin{array}{rl}& p=\frac{\pi }{2}\\ & F=\frac{1}{p}=\frac{2}{\pi }\end{array}$

(2)- اختبار موضوعي

1) ضع إشارة + أو – في المستطيلات التالية لتحصل على عبارة صحيحة:

1. $\cos {({20}^{\circ }+50)}^{\circ }=\cos {20}^{\circ }\cos {50}^{\circ }\mathbf{-}\sin {20}^{\circ }\sin {50}^{\circ }$
2. $\tan (3\mathrm{A}-2\mathrm{B})=\tan 3\mathrm{A}\mathbf{-}\tan 2\mathrm{B}/1\mathbf{+}\tan 3\mathrm{Atan}2\mathrm{B}$
3. $\sin (80\mathbf{-}10)°=\sin 80\cos {10}^{\circ }-\cos {80}^{\circ }\sin {10}^{\circ }$

2) أكمل ما يأتي لتحصل على عبارة صحيحة

1. $\sin ({40}^{\circ }+{180}^{\circ })=\sin {40}^{\circ }\mathbf{cos}\mathbf{}{\mathbf{180}}^{\mathbf{\circ }}+\mathbf{cos}\mathbf{}{\mathbf{40}}^{\mathbf{\circ }}\sin {180}^{\circ }$
2. $2\sin \mathrm{\Pi }/3\cos \mathrm{\Pi }/3=\sin \frac{\mathbf{2}\mathbf{\pi }}{\mathbf{3}}$
3. ${\cos }^2{15}^{\circ }-{\sin }^2{15}^{\circ }=\cos \mathbf{ }{\mathbf{30}}^{\mathbf{\circ }}$

(3)- عين العبارات الصحيحة والعبارات الخاطئة فيما يأتي:

1. $\sin 6x=2\sin 3x$ (خطأ)
2. $\sin 15\cos {15}^{\circ }=\sin 30$ (خطأ)
3. $\cos {80}^{\circ }={\cos }^2{40}^{\circ }-{\sin }^2{40}^{\circ }$ (صح)
4. مجموعة حل المعادلة $2\cos x+3=0$ هي $\mathrm{\varnothing }$ (صح)

(4)- اختر من القائمة A ما يناسبها من القائمة B

1) $\cos 4\mathrm{Acos}\mathrm{A}-\sin 4\mathrm{Asin}\mathrm{A}=$

$\cos 5\mathrm{A}$

2) $\sin \mathrm{Acos}4\mathrm{A}-\sin 4\mathrm{Acos}\mathrm{A}=$

$\sin (-3\mathrm{A})$

3) $\sin 4\mathrm{Acos}\mathrm{A}+\cos 4\mathrm{Asin}\mathrm{A}=$

$\sin 5\mathrm{A}$

(5)- اختبار مقالي:

1) إذا كان $\frac{3\mathrm{\Pi }}{2}<x<2\mathrm{\Pi }$ وكانت $\cos \mathrm{x}=\frac{2}{3}$ فأوجد قيمة كل من: cot x, sec x, cscx

$\begin{array}{rl}& \mathrm{a}=\sqrt{5}\\ & \cot \mathrm{x}=\frac{1}{\tan \mathrm{x}}=\frac{1}{-\frac{\sqrt{5}}{2}}=\frac{-2}{\sqrt{5}}\\ & \sec \mathrm{x}=\frac{1}{\cos \mathrm{x}}=\frac{3}{2}\\ & \csc \mathrm{x}=\frac{1}{\sin \mathrm{x}}=-\frac{3}{\sqrt{5}}\end{array}$

2) إذا كان $\frac{3\mathrm{\Pi }}{2}<x<2\mathrm{\Pi }$ وكانت $\cos \mathrm{x}=\frac{3}{5}$ فأوجد قيمة كل من: $\cos 2\mathrm{x} , \sin 2\mathrm{x} , \tan 2\mathrm{x}$

$$\begin{gathered} {\sin }^2\mathrm{x}=1-{\cos }^2\mathrm{x}=1-\frac{9}{25}=\frac{25}{25}-\frac{9}{25}=\frac{16}{25}\to \sin \mathrm{x}=-\frac{4}{5} \\ \cos 2\mathrm{x}=2{\cos }^2\mathrm{x}-1=2\times \frac{9}{25}-1=\frac{18-25}{25}=-\frac{7}{25} \end{gathered}$$

$\begin{array}{rl}& \sin 2\mathrm{x}=2\sin \mathrm{xcos}\mathrm{x}=2\times \frac{-4}{5}\times \frac{3}{5}=\frac{-24}{25}\\ & \tan \mathrm{x}=\frac{\sin \mathrm{x}}{\cos \mathrm{x}}=-\frac{4}{5}÷\frac{3}{5}=-\frac{4}{5}\times \frac{5}{3}=\frac{-4}{3}\\ & \tan 2\mathrm{x}=\frac{2\tan \mathrm{x}}{1-{\tan }^2\mathrm{x}}=\frac{2\times \frac{-4}{3}}{1-{(-\frac{4}{3})}^2}=\frac{-\frac{8}{3}}{1-\frac{16}{9}}=\frac{-\frac{3}{7}}{-\frac{7}{9}}=\frac{8}{3}÷\frac{7}{9}=\frac{8}{3}\times \frac{9}{7}=\frac{24}{7}\end{array}$

3) بدون استخدام الحاسبة أوجد قيمة:

أ- $\sin \mathrm{\Pi }/8\cos \mathrm{\Pi }/8$

$\begin{array}{rl}\sin \frac{\mathrm{\pi }}{8}\cos \frac{\mathrm{\pi }}{8}& =1\times \sin \frac{\mathrm{\pi }}{8}\cos \frac{\mathrm{\pi }}{8}\\ & =\frac{2}{2}\sin \frac{\mathrm{\pi }}{8}\cos \frac{\mathrm{\pi }}{8}=\frac{1}{2}(2\sin \frac{\mathrm{\pi }}{8}\cos \frac{\mathrm{\pi }}{8})\\ & \begin{array}{rl}& =\frac{1}{2}(\sin 2\left(\frac{\pi }{8}\right)=\frac{1}{2}\sin \frac{2\pi }{8}=\frac{1}{2}\sin \frac{\pi }{4}\\ & =\frac{1}{2}\sin {45}^{\circ }=\frac{1}{2}\cdot \frac{1}{\sqrt{2}}=\frac{1}{2}\cdot \frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{4}\end{array}\end{array}$

ب- ${\cos }^2\frac{\mathrm{\Pi }}{12}-{\sin }^2\frac{\mathrm{\Pi }}{12}$

$\begin{array}{rl}{\cos }^2\frac{\mathrm{\pi }}{12}& =\frac{1+\cos 2\left(\frac{\mathrm{\pi }}{12}\right)}{2}=\frac{1+\cos \frac{2\mathrm{\pi }}{12}}{2}\\ & =\frac{1+\cos \frac{\mathrm{\pi }}{6}}{2}=1+\frac{1}{2}=\frac{1\frac{1}{2}}{2}=\frac{3}{2}=\frac{3}{4}\end{array}$

4) أثبت صحة المتطابقة الآتية:

${\cos }^4\mathrm{x}-{\sin }^4\mathrm{x}=\cos 2\mathrm{x}$

$\begin{array}{rl}\text{L.S=}{\cos }^4\mathrm{x}-{\sin }^4\mathrm{x}& =({\cos }^2\mathrm{x}-{\sin }^2\mathrm{x})({\cos }^2\mathrm{x}+{\sin }^2\mathrm{x})\\ & =({\cos }^2\mathrm{x}-{\sin }^2\mathrm{x})\times 1\\ & =\cos 2\mathrm{x}=\mathrm{R}\cdot \mathrm{S}\end{array}$