The Collectors

Bài 6.18 trang 191 SBT đại số 10

Câu hỏi: Không dùng bảng số và máy tính, chứng minh rằng
a) $\sin 20^{\circ}+2 \sin 40^{\circ}-\sin 100^{\circ}=\sin 40^{\circ}$
b) $\frac{\sin \left(45^{\circ}+\alpha\right)-\cos \left(45^{\circ}+\alpha\right)}{\sin \left(45^{\circ}+\alpha\right)+\cos \left(45^{\circ}+\alpha\right)}=\tan \alpha$
c) $\frac{3 \cot ^{2} 15^{\circ}-1}{3-\cot ^{2} 15^{\circ}}=-\cot 15^{\circ}$
d) $\sin 200^{\circ} \sin 310^{\circ}+\cos 340^{\circ} \cos 50^{\circ}=\frac{\sqrt{3}}{2}$
a)
$\begin{array}{l}
\sin {20^\circ } + 2\sin {40^\circ } - \sin {100^\circ }\\
= \left( {\sin {{20}^\circ } - \sin {{100}^\circ }} \right) + 2\sin {40^\circ }\\
= 2\cos {60^\circ }\sin \left( { - {{40}^\circ }} \right) + 2\sin {40^\circ }\\
= - \sin {40^\circ } + 2\sin {40^\circ } = \sin {40^\circ }
\end{array}$
b)
$\begin{aligned} \frac{\sin \left(45^{\circ}+\alpha\right)-\cos \left(45^{\circ}+\alpha\right)}{\sin \left(45^{\circ}+\alpha\right)+\cos \left(45^{\circ}+\alpha\right)} &=\frac{\sin \left(45^{\circ}+\alpha\right)-\sin \left(45^{\circ}-\alpha\right)}{\sin \left(45^{\circ}+\alpha\right)+\sin \left(45^{\circ}-\alpha\right)} \\ &=\frac{2 \cos 45^{\circ} \sin \alpha}{2 \sin 45^{\circ} \cos \alpha}=\frac{\sqrt{2} \sin \alpha}{\sqrt{2} \cos \alpha}=\tan \alpha \end{aligned}$
c)
$\begin{aligned} A &=\frac{3 \cot ^{2} 15^{\circ}-1}{3-\cot ^{2} 15^{\circ}}=\frac{\cot ^{2} 30^{\circ} \cot ^{2} 15^{n}-1}{\cot ^{2} 30^{\circ}-\cot ^{2} 15^{\circ}} \\ &=\frac{\cot 30^{\circ} \cot 15^{\circ}+1}{\cot 30^{\circ}-\cot 15^{\circ}} \cdot \frac{\cot 30^{\circ} \cot 15^{\circ}-1}{\cot 30^{\circ}+\cot 15^{\circ}} \end{aligned}$
Mặt khác ta có: $\cot (\alpha+\beta)=\frac{\cos (\alpha+\beta)}{\sin (\alpha+\beta)}=\frac{\cos \alpha \cos \beta-\sin \alpha \sin \beta}{\sin \alpha \cos \beta+\cos \alpha \sin \beta}$
Chia cả tử và mẫu của biểu thức cho $\sin \alpha \sin \beta$ ta được
$
\cot (\alpha+\beta)=\frac{\cot \alpha \cot \beta-1}{\cot \alpha+\cot \beta}
$
Tương tự ta có:
$
\cot (\alpha-\beta)=\frac{\cot \alpha \cot \beta+1}{\cot \beta-\cot \alpha}
$
Do đó:
$
A=\cot \left(15^{\circ}-30^{\circ}\right) \cot \left(15^{\circ}+30^{\circ}\right)=-\cot 15^{\circ}
$
d)
$\sin 200^{\circ} \sin 310^{\circ}+\cos 340^{\circ} \cos 50^{\circ}$
$
\begin{aligned} &=\sin \left(180^{\circ}+20^{\circ}\right) \sin \left(360^{\circ}-50^{\circ}\right)+\cos \left(360^{\circ}-20^{\circ}\right) \cos 50^{\circ} \\ &=\left(-\sin 20^{\circ}\right)\left(-\sin 50^{\circ}\right)+\cos 20^{\circ} \cos 50^{\circ} \end{aligned}
$
$=\cos 50^{\circ} \cos 20^{\circ}+\sin 50^{\circ} \sin 20^{\circ}$
$=\cos \left(50^{\circ}-20^{\circ}\right)=\frac{\sqrt{3}}{2}$
 

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