Quay lại đáp án

Áp dụng công thức $sin^4x=\frac{cos4x-4cos2x+3}8$Thế vào A, ta có $8A = -4cos\frac{\pi}8-4cos\frac{3\pi}8-4cos\frac{5\pi}8-4cos\frac{7\pi}8+12$$\Leftrightarrow -2A = cos\frac{\pi}8 +cos\frac{7\pi}8+cos\frac{5\pi}8+cos\frac{3\pi}8+3$$\Leftrightarrow -2A=(2cos\pi.cos\frac{3\pi}4+2cos\pi.cos\frac{\pi}4)+3 $$\Leftrightarrow -2A=(2cos\pi.2.cos\pi.cos\frac{\pi}2)+3$$\Leftrightarrow -2A=0+3 \Rightarrow A=-\frac{3}2$

Áp dụng công thức $ sin^4x=\frac{cos4x-4cos2x+3}8 $ Thế vào A, ta có $8A = -4cos\frac{\pi}8-4cos\frac{3\pi}8-4cos\frac{5\pi}8-4cos\frac{7\pi}8+12$$\Leftrightarrow -2A = cos\frac{\pi}8 +cos\frac{7\pi}8+cos\frac{5\pi}8+cos\frac{3\pi}8+3$$\Leftrightarrow -2A=(2cos\pi.cos\frac{3\pi}4+2cos\pi.cos\frac{\pi}4)+3 $$\Leftrightarrow -2A=(2cos\pi.2.cos\pi.cos\frac{\pi}2)+3$$\Leftrightarrow -2A=0+3 \Rightarrow A=-\frac{3}2$

Áp dụng công thức $sin^4x=\frac{cos4x-4cos2x+3}8$Thế vào A, ta có $8A = -4cos\frac{\pi}8-4cos\frac{3\pi}8-4cos\frac{5\pi}8-4cos\frac{7\pi}8+12$$\Leftrightarrow -2A = cos\frac{\pi}8 +cos\frac{7\pi}8+cos\frac{5\pi}8+cos\frac{3\pi}8+3$$\Leftrightarrow -2A=(2cos\pi.cos\frac{3\pi}4+2cos\pi.cos\frac{\pi}4)+3 $$\Leftrightarrow -2A=(2cos\pi.2.cos\pi.cos\frac{\pi}2)+3$$\Leftrightarrow -2A=0+3 \Rightarrow A=-\frac{3}2$

Áp dụng công thức $sin^4x=\frac{cos4x-4cos2x+3}8$Thế vào A, ta có $8A = -4cos\frac{\pi}8-4cos\frac{3\pi}8-4cos\frac{5\pi}8-4cos\frac{7\pi}8+12$$\Leftrightarrow -2A = cos\frac{\pi}8 +cos\frac{7\pi}8+cos\frac{5\pi}8+cos\frac{3\pi}8+3$$\Leftrightarrow -2A=(2cos\pi.cos\frac{3\pi}4+2cos\pi.cos\frac{\pi}4)+3 $$\Leftrightarrow -2A=(2cos\pi.2.cos\pi.cos\frac{\pi}2)+3$$\Leftrightarrow -2A=0+3 \Rightarrow A=-\frac{3}2$

Áp dụng công thức $sin^4x=\frac{cos4x-4cos2x+3}8$Thế vào A, ta có $8A = -4cos\frac{\pi}8-4cos\frac{3\pi}8-4cos\frac{5\pi}8-4cos\frac{7\pi}8+12$$\Leftrightarrow -2A = cos\frac{\pi}8 +cos\frac{7\pi}8+cos\frac{5\pi}8+cos\frac{3\pi}8+3$$\Leftrightarrow -2A=(2cos\pi.cos\frac{3\pi}4+2cos\pi.cos\frac{\pi}4)+3 $$\Leftrightarrow -2A=(2cos\pi.2.cos\pi.cos\frac{\pi}2)+3$$\Leftrightarrow -2A=0+3 \Rightarrow A=-\frac{3}2$

Áp dụng công thức $sin^4x=\frac{cos4x-4cos2x+3}8$Thế vào A, ta có $8A = -4cos\frac{\pi}8-4cos\frac{3\pi}8-4cos\frac{5\pi}8-4cos\frac{7\pi}8+12$$\Leftrightarrow -2A = cos\frac{\pi}8 +cos\frac{7\pi}8+cos\frac{5\pi}8+cos\frac{3\pi}8+3$$\Leftrightarrow -2A=(2cos\pi.cos\frac{3\pi}4+2cos\pi.cos\frac{\pi}4)+3 $$\Leftrightarrow -2A=(2cos\pi.2.cos\pi.cos\frac{\pi}2)+3$$\Leftrightarrow -2A=0+3 \Rightarrow A=-\frac{3}2$