حل أسئلة مراجعة الدرس الثالث: المتطابقات المثلثية لمجموع زاويتين والفرق بينهما

دون استعمال الآلة الحاسبة، أوجد القيمة الدقيقة لكل مما يأتي:

24) cos (-135°)

$\begin{array}{rl}& \cos (-135)=\cos (45-180)\\ & =\cos 45\cos 180+\sin 45\sin 180\\ & =\frac{\sqrt{2}}{2}\times 1+\frac{\sqrt{2}}{2}\times 0\\ & =\frac{\sqrt{2}}{2}\end{array}$

25) cos 15°

$\begin{array}{rl}& \cos \left(15\right)=\cos (45-30)\\ & =\cos 45\cos 30+\sin 45\sin 30\\ & =\frac{\sqrt{2}}{2}\times \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times \frac{1}{2}\\ & =\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}\\ & =\frac{\sqrt{6}+\sqrt{2}}{4}\end{array}$

26) sin 210°

$\begin{array}{rl}& \sin \left(210\right)=\sin (180+30)\\ & =\sin 180\cos 30+\cos 180\sin 30\\ & =0\times \frac{\sqrt{3}}{2}+1\times \frac{1}{2}\\ & =\frac{1}{2}\end{array}$

27) sin 105°

$\begin{array}{rl}& \sin \left(105\right)=\sin (60+45)\\ & =\sin 60\cos 45+\cos 60\sin 45\\ & =\frac{\sqrt{3}}{2}\times \frac{\sqrt{2}}{2}+\frac{1}{2}\times \frac{\sqrt{2}}{2}\\ & =\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}\\ & =\frac{\sqrt{6}+\sqrt{2}}{4}\end{array}$

28) tan 75°

$\begin{array}{rl}& tan\left(75\right)=tan(45+30)\\ & =\frac{tan45+tan30}{1-\tan 45tan30}\\ & =\frac{1+\sqrt{3}}{1-\sqrt{3}}\end{array}$

29) cos 105°

$\begin{array}{rl}& \cos \left(105\right)=\cos (60+45)\\ & =\cos 60\cos 45-\sin 60\sin 45\\ & =\frac{1}{2}\times \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{2}}{2}\\ & =\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\\ & =\frac{\sqrt{6}+\sqrt{2}}{4}\end{array}$

أثبت صحة كل من المتطابقات الآتية:

30) sin (θ + 90 ) = cos θ

$\begin{array}{rl}& sin(\theta +90)\\ & =sin\theta cos90+cos\theta sin90\\ & =sin\theta \times 0+cos\theta \times 1\\ & =cos\theta \end{array}$

31) $\sin \left(\frac{3\pi }{2}-\theta \right)=-\cos \theta$

$\begin{array}{rl}& =sin\frac{3\pi }{2}cos\theta +cos\frac{3\pi }{2}sin\theta \\ & =-1\times cos\theta -0\times sin\theta \\ & =-cos\theta \end{array}$

32) tan (θ - π) = tan θ

$\begin{array}{rl}& tan(\theta -\pi )\\ & =\frac{tan\theta -tan\pi }{\mathbf{1}\mathbf{+}tan\theta tan\pi }\\ & =\frac{tan\theta -\mathbf{0}}{\mathbf{1}\mathbf{+}\mathbf{0}}\\ & =tan\theta \end{array}$