Question

已知d∈(π/2,π),且tan(a+)π/4)=1/7,那么sina+cosa的值为-1/5为什么？

Answer 1

Answer 2

tan(a+π/4)=1/7
﹙tana+tanπ/4)／﹙1-tanatanπ/4﹚=1/7
﹙tana+1)／﹙1-tana﹚=1/7
tana+1=1/7﹙1-tana﹚
sina+cosa=1/7（cosa-sina）
7sina+7cosa=cosa-sina
4sina+3cosa=0
16sin²a+16co²sa=16=9cos²a+16cos²a=25cos²a
∴cos²a=16／25
∵a∈(π/2,π),
∴cosa=﹣4／5
∴sina＝√﹙1-16／25﹚＝3／5
∴sina+cosa的值为-1/5
或
sin(a+π/4)／cos(a+π/4)=1/7
sin(a+π/4)=1/7cos(a+π/4﹚
7sin(a+π/4)=cos(a+π/4﹚
49sin²(a+π/4)=cos²(a+π/4﹚=1-sin²(a+π/4)
sin²(a+π/4)=1／50
∵a∈(π/2,π),
∴a+π/4∈(3π/4,5π/4),
∵tan(a+π/4)=1/7＞0
∴a+π/4∈(π,5π/4),
∴sin(a+π/4)=﹣√2／10
∴√2／2﹙sina+cosa)=﹣√2／10
∴sina+cosa)=﹣1／5

Answer 3

d∈(π/2,π)还是a∈(π/2,π)?