Bài 113047

Question

Tìm giá trị các biểu thức
$1.Q=\tan^2 \frac{\pi}{12}+\tan^2 \frac{3\pi}{12}+\tan^2 \frac{5\pi}{12}$
$2.R=\sin^4 \frac{\pi}{16}+\sin^4 \frac{3\pi}{16}+\sin^4 \frac{5\pi}{16}+\sin^4 \frac{7\pi}{16}$

score 1 · Answer 1

1.Ta có:
$* \tan^2 \frac{\pi}{12}=\frac{2\sin^2 \frac{\pi}{12} }{2\cos^2 \frac{\pi}{12} }=\frac{1-\cos \frac{\pi}{6} }{1+\cos \frac{\pi}{6} }=\frac{1-\frac{\sqrt{3}}{2} }{1+\frac{\sqrt{3}}{2} }=(2-\sqrt{3})^2=7-4 \sqrt{3}$
$* \tan^2 \frac{3\pi}{12}=\tan^2 \frac{\pi}{4}=1$
$* \tan^2 \frac{5\pi}{12}=\cot^2(\frac{\pi}{2}-\frac{5\pi}{12})=\cot^2 \frac{\pi}{12}=\frac{1}{7-4 \sqrt{3} }=7+4 \sqrt{3}$
Suy ra $\tan^2 \frac{\pi}{12}+\tan^2 \frac{3\pi}{12}+\tan^2 \frac{5\pi}{12}=7-4 \sqrt{3}+1+7+4 \sqrt{3}=15$. Vậy $Q=15$
2.Ta có: $2\sin^2 a=1-\cos 2a$
$\Rightarrow 4\sin^2 a=(1-\cos 2a)^2=1-2\cos 2a+\cos^22a=1-\cos 2a$
$\Rightarrow 8\sin^4 a=2-4\cos 2a+(1+\cos 4a)=3-4\cos 2a+\cos 4a$
Tóm lại $8\sin^4 a=3-4\cos 2a+\cos 4a   (1)$
Áp dụng $(1)$ lần lượt với $a=\frac{\pi}{16};\frac{3\pi}{16},\frac{5\pi}{16},\frac{7\pi}{16}$
$* 8\sin^4 \frac{\pi}{16}=3-4\cos \frac{\pi}{8}+\cos \frac{\pi}{4}$
$* 8\sin^4 \frac{3\pi}{16}=3-4\cos \frac{3\pi}{8}+\cos \frac{3\pi}{4}$
$* 8\sin^4 \frac{5\pi}{16}=3-4\cos \frac{5\pi}{8}+\cos \frac{5\pi}{4}$
$* 8\sin^4 \frac{7\pi}{16}=3-4\cos \frac{7\pi}{8}+\cos \frac{7\pi}{4}$
Cộng theo từng vế có:
$8R=12-4(\cos \frac{\pi}{8}+\cos \frac{7\pi}{8})-4(\cos \frac{3\pi}{8}+\cos \frac{5\pi}{8})\\+(\cos \frac{\pi}{4}+\cos \frac{3\pi}{4})+(\cos \frac{5\pi}{4}+\cos \frac{7\pi}{4})$
Để ý: $\cos \frac{7\pi}{8}=-\cos (\pi-\frac{7\pi}{8})=-\cos \frac{\pi}{8} \Leftrightarrow \cos \frac{\pi}{8}+\cos \frac{7\pi}{8}=0$
     $\cos \frac{7\pi}{4}=-\cos (\pi-\frac{7\pi}{4})=-\cos \frac{5\pi}{4} \Leftrightarrow \cos \frac{3\pi}{8}+\cos \frac{5\pi}{8}=0      (2)$
Thay $(2)$ vào $(1)$ có $8R=12 \Leftrightarrow R=\frac{3}{2}$

tại sao ở ý b, khi bình phương 2 vế lên rồi sin vẫn giữ nguyên ở mũ 2 vậy?

Bài 113047

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Bài 113047

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