Câu hỏi của Jackson Roy

Question

1. Biểu thức A = $\frac{1}{2\sin10}-2\sin70$ có gái trị bằng bao nhiêu ?

2. Tích số cos10.cos30.cos50.cos70 = ?

3. Tích số $cos\frac{\pi}{7}.cos\frac{4\pi}{7}.cos\frac{5\pi}{7}$ = ?

4. Tính A...

Nguyễn Việt Lâm · Accepted Answer

Câu 3:

$A=cos\frac{\pi}{7}.cos\frac{5\pi}{7}.cos\frac{4\pi}{7}=cos\frac{\pi}{7}.cos\left(\pi-\frac{2\pi}{7}\right).cos\frac{4\pi}{7}$

$A=-cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}$

$\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.2sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}$

$\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.sin\frac{2\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}$

$\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{4}sin\frac{4\pi}{7}.cos\frac{4\pi}{7}$

$\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{8}sin\frac{8\pi}{7}=-\frac{1}{8}sin\left(\pi+\frac{\pi}{7}\right)=\frac{1}{8}sin\frac{\pi}{7}$

$\Rightarrow A=\frac{1}{8}$

Câu 4:

Đầu tiên ta chứng minh công thức:

$tana+tanb=\frac{sina}{cosa}+\frac{sinb}{cosb}=\frac{sina.cosb+cosa.sinb}{cosa.cosb}=\frac{sin\left(a+b\right)}{cosa.cosb}$

Áp dụng để biến đổi tử số:

$tan30+tan60+tan40+tan50=\frac{sin90}{cos30.cos60}+\frac{sin90}{cos40.cos50}=\frac{1}{cos30.cos60}+\frac{1}{cos40.cos50}$

$=\frac{2}{cos90+cos30}+\frac{2}{cos90+cos10}=\frac{2}{cos30}+\frac{2}{cos10}=2\left(\frac{cos30+cos10}{cos30.cos10}\right)$

$=2\left(\frac{2cos20.cos10}{cos30.cos10}\right)=\frac{4.cos20}{cos30}=\frac{8\sqrt{3}}{3}.cos20$

$\Rightarrow A=\frac{\frac{8\sqrt{3}}{3}cos20}{cos20}=\frac{8\sqrt{3}}{3}$

Câu 5:

$cos54.cos4-cos36.cos86=cos54.cos4-cos\left(90-54\right).cos\left(90-4\right)$

$=cos54.cos4-sin54.sin4=cos\left(54+4\right)=cos58$Câu trả lời: Câu 3:

$A=cos\frac{\pi}{7}.cos\frac{5\pi}{7}.cos\frac{4\pi}{7}=cos\frac{\pi}{7}.cos\left(\pi-\frac{2\pi}{7}\right).cos\frac{4\pi}{7}$

$A=-cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}$

$\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.2sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}$

$\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.sin\frac{2\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}$

$\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{4}sin\frac{4\pi}{7}.cos\frac{4\pi}{7}$

$\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{8}sin\frac{8\pi}{7}=-\frac{1}{8}sin\left(\pi+\frac{\pi}{7}\right)=\frac{1}{8}sin\frac{\pi}{7}$

$\Rightarrow A=\frac{1}{8}$

Câu 4:

Đầu tiên ta chứng minh công thức:

$tana+tanb=\frac{sina}{cosa}+\frac{sinb}{cosb}=\frac{sina.cosb+cosa.sinb}{cosa.cosb}=\frac{sin\left(a+b\right)}{cosa.cosb}$

Áp dụng để biến đổi tử số:

$tan30+tan60+tan40+tan50=\frac{sin90}{cos30.cos60}+\frac{sin90}{cos40.cos50}=\frac{1}{cos30.cos60}+\frac{1}{cos40.cos50}$

$=\frac{2}{cos90+cos30}+\frac{2}{cos90+cos10}=\frac{2}{cos30}+\frac{2}{cos10}=2\left(\frac{cos30+cos10}{cos30.cos10}\right)$

$=2\left(\frac{2cos20.cos10}{cos30.cos10}\right)=\frac{4.cos20}{cos30}=\frac{8\sqrt{3}}{3}.cos20$

$\Rightarrow A=\frac{\frac{8\sqrt{3}}{3}cos20}{cos20}=\frac{8\sqrt{3}}{3}$

Câu 5:

$cos54.cos4-cos36.cos86=cos54.cos4-cos\left(90-54\right).cos\left(90-4\right)$

$=cos54.cos4-sin54.sin4=cos\left(54+4\right)=cos58$

Nguyễn Việt Lâm · Answer

Câu 1:

$A=\frac{1}{2sin10}-2sin70=\frac{1-4sin10.sin70}{2sin10}=\frac{1+2\left(cos80-cos60\right)}{2sin10}$

$=\frac{1+2cos80-1}{2sin10}=\frac{2cos80}{2sin10}=\frac{sin10}{sin10}=1$

Câu 2:

$cos10.cos30.cos50.cos70=cos10.cos30.\frac{1}{2}\left(cos120+cos20\right)$

$=\frac{1}{2}cos30\left(cos10.cos120+cos10.cos20\right)$

$=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}\left(cos30+cos10\right)\right)$

$=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}cos30+\frac{1}{2}cos10\right)$

$=\frac{1}{2}.\frac{\sqrt{3}}{2}\left(-\frac{1}{2}cos10+\frac{1}{2}\frac{\sqrt{3}}{2}+\frac{1}{2}cos10\right)$

$=\frac{3}{16}$Câu trả lời: Câu 1: