أثبت صحة كل من المتطابقات الآتية:

$\cot \theta +\sec \theta =\frac{{\cos }^2\theta +\sin \theta }{\sin \theta \cos \theta }$

$\begin{array}{c}\frac{co{s}^2\theta +sin\theta }{sin\theta cos\theta }=\frac{co{s}^2\theta }{sin\theta cos\theta }+\frac{sin\theta }{sin\theta cos\theta }\\ =\frac{cos\theta }{sin\theta }+\frac{1}{cos\theta }=cot\theta +sec\theta \end{array}$

${\sin }^2\theta +{\tan }^2\theta =(1-{\cos }^2\theta )+\frac{{\sec }^2\theta }{{\csc }^2\theta }$

$(1-co{s}^2\theta )+\frac{se{c}^2\theta }{cs{c}^2\theta }=si{n}^2\theta +\frac{\frac{1}{co{s}^2\theta }}{\frac{1}{si{n}^2\theta }}=si{n}^2\theta +ta{n}^2\theta$

$(\sin \theta -\cos \theta {)}^2=1-2\sin \theta \cos \theta$

$\begin{array}{rl}(sin\theta -cos\theta {)}^2& =si{n}^2\theta +co{s}^2\theta -2sin\theta cos\theta \\ & =1-2sin\theta cos\theta \end{array}$

حل أسئلة مراجعة تراكمية

أثبت صحة كل من المتطابقات الآتية:

33) $\cot \theta +\sec \theta =\frac{{\cos }^2\theta +\sin \theta }{\sin \theta \cos \theta }$

$\begin{array}{c}\frac{co{s}^2\theta +sin\theta }{sin\theta cos\theta }=\frac{co{s}^2\theta }{sin\theta cos\theta }+\frac{sin\theta }{sin\theta cos\theta }\\ =\frac{cos\theta }{sin\theta }+\frac{1}{cos\theta }=cot\theta +sec\theta \end{array}$

34) ${\sin }^2\theta +{\tan }^2\theta =(1-{\cos }^2\theta )+\frac{{\sec }^2\theta }{{\csc }^2\theta }$

$(1-co{s}^2\theta )+\frac{se{c}^2\theta }{cs{c}^2\theta }=si{n}^2\theta +\frac{\frac{1}{co{s}^2\theta }}{\frac{1}{si{n}^2\theta }}=si{n}^2\theta +ta{n}^2\theta$

35) $(\sin \theta -\cos \theta {)}^2=1-2\sin \theta \cos \theta$

$\begin{array}{rl}(sin\theta -cos\theta {)}^2& =si{n}^2\theta +co{s}^2\theta -2sin\theta cos\theta \\ & =1-2sin\theta cos\theta \end{array}$

أوجد القيمة الدقيقة لكل مما يأتي:

36) sin 135°

$\sin 135=\frac{\sqrt{2}}{2}$

37) cos 105°

$\cos 105=\frac{\sqrt{2}-\sqrt{6}}{4}$

38) sin 285°

$\sin 285=\frac{-\sqrt{6}-\sqrt{2}}{4}$

39) cos 210°

$\cos \left(210\right)=-\frac{\sqrt{3}+2}{4}$

40) sin (-240°)

$\sin (-240)=\frac{\sqrt{3}}{2}$

41) cos (-120°)

$\cos (-120)=-\frac{1}{2}$

42) cos 78° cos 18° + sin 78° sin 18°.

$\cos 78\cos 18+\sin 78\sin 18=\cos (78-18)=\cos 60=\frac{1}{2}$