高一数学题(1)化简:1-tanα·sin(α-2π)·sin(π/2+α)(2)若角α=-17/4π,试求(1)式的值
(1)化简:1-tanα·sin(α-2π)·sin(π/2+α)
(2)若角α=-17/4π,试求(1)式的值
人气:195 ℃ 时间:2020-05-27 14:12:53
解答
(1)1-tanα·sin(α-2π)·sin(π/2+α)=1+tanα·sinα·cosα=1+sin²a
(2)原式=1+sin²(-17/4π)=1+sin²π/4=3/2
推荐
- 已知2sinβ=sin(2α+β),求tan(α+β)/tanα的值
- 高一数学月考题.已知sin[π+α]=-1/2,求sinα,cosα,tanα的值
- 已知tanα=1/2,则((1/4)sin^2)
- sin(π/2+a)=-1/2求tan(a-2π)这是高一数学两角和与差的余弦
- Cosα=1/4,求Sinα,tanα
- 已知X1 ,X2 是关于一元二次方程4X^2 + 4(M+1)X + M^2 =0的两个非零实数根问X1,X2能否同号?若能求出M 的取
- 圆形-五角=三角 - + * 五角+圆形=正方 = = = 三角*正方=8 求圆形,三角,五角,正方=?
- The reason ________ I'm phoning to you is to tell you about a farewell party on Saturday.
猜你喜欢