Câu hỏi của Lan Hương

Question

tính nhanh

A=$sin^242^o+sin^243^o+sin^244^o+sin^245^0+sin^246^o+sin^247^o+sin^248^o$

B=$\cos^215^o-cos^225^o+cos^235^o-cos^245^o+cos^255^o-cos^265^o+cos^275^o$

Hoàng Tử Hà · Accepted Answer

$ADCT:\sin^2\alpha+\cos^2\alpha=1$

$A=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin45^0$

$A=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\frac{\sqrt{2}}{2}$

$A=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}$

Câu b lm tương tựCâu trả lời: $ADCT:\sin^2\alpha+\cos^2\alpha=1$

$A=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin45^0$

$A=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\frac{\sqrt{2}}{2}$

$A=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}$

Câu b lm tương tự