Tính các giá trị lượng giác của cung α\alphaα, biết :
a) sinα=0,6\sin\alpha=0,6sinα=0,6 khi 0<α<π20< \alpha< \dfrac{\pi}{2}0<α<2π
b) cosα=−0,7\cos\alpha=-0,7cosα=−0,7 khi π2<α<π\dfrac{\pi}{2}< \alpha< \pi2π<α<π
c) tanα=2\tan\alpha=2tanα=2 khi π<α<3π2\pi< \alpha< \dfrac{3\pi}{2}π<α<23π
d) cotα=−3\cot\alpha=-3cotα=−3 khi 3π2<α<2π\dfrac{3\pi}{2}< \alpha< 2\pi23π<α<2π
Do π2<α<π\dfrac{\pi}{2}< \alpha< \pi2π<α<π nên sinα>0;tanα<0;cotα<0sin\alpha>0;tan\alpha< 0;cot\alpha< 0sinα>0;tanα<0;cotα<0. sinα=1−cos2α=5110sin\alpha=\sqrt{1-cos^2\alpha}=\dfrac{\sqrt{51}}{10}sinα=1−cos2α=1051. tanα=sinαcosα=5110:(−0,7)=−517tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\sqrt{51}}{10}:\left(-0,7\right)=-\dfrac{\sqrt{51}}{7}tanα=cosαsinα=1051:(−0,7)=−751. cotα=1tanα=−751cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-7}{\sqrt{51}}cotα=tanα1=51−7.
Bài 27 (SBT trang 195)
Hãy xác định dấu của các tích (không dùng bảng số và máy tính)
a) sin1100cos1300tan300cot3200\sin110^0\cos130^0\tan30^0\cot320^0sin1100cos1300tan300cot3200
b) sin(−500)tan1700cos(−910)sin5300\sin\left(-50^0\right)\tan170^0\cos\left(-91^0\right)\sin530^0sin(−500)tan1700cos(−910)sin5300
Bài 26 (SBT trang 195)
Hãy viết theo thứ tự tăng dần các giá trị sau ( không dùng bảng số và máy tính) :
a) sin400,sin900,sin2200,sin100\sin40^0,\sin90^0,\sin220^0,\sin10^0sin400,sin900,sin2200,sin100
b) cos150,cos00,cos900,cos1380\cos15^0,\cos0^0,\cos90^0,\cos138^0cos150,cos00,cos900,cos1380
Tìm số nguyên n thỏa mãn từng điều kiện sau:
a) (n+1)(n+3)=0
b) (|n|+2)(n²-1)=0
Cho x,y,z là ba số không âm thỏa mãn: x + y + z = 1. Chứng minh rằng:
x1+x2+y1+y2+z1+z2≤910\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\le\frac{9}{10}1+x2x+1+y2y+1+z2z≤109
giải bất phương trình sau:
∣5x+2∣<∣10x−1∣\left|\frac{5}{x+2}\right|< \left|\frac{10}{x-1}\right|∣∣∣∣x+25∣∣∣∣<∣∣∣∣x−110∣∣∣∣
(nói rõ cách giải giúp mình, vì có nhiều chỗ mình chưa hiểu)
Giải các Phtr sau ☘Tiểu Tuyết☘
a) 26x3 - 12x2 + 13x = 6
b) (x + 1)3 - ( x - 1)3 = 6 ( x2 + x + 1)
c) 3x−1x−1−2x+5x+3−8x2+2x−3=1\frac{3x-1}{x-1}-\frac{2x+5}{x+3}-\frac{8}{x^2+2x-3}=1x−13x−1−x+32x+5−x2+2x−38=1
Tính nhanh
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tim so co ba chu so, biet rang khi viet them chu so 1 vao ben trai so do thi duoc so co bon chu so gap 9lan so phai tim.
bai giai
2x−1\sqrt{2x-1}2x−1 = xxx + 1
Chứng minh bất đẳng thức: a2+b2+c2+34≥a+b+ca^{^{ }2}+b^2+c^2+\dfrac{3}{4}\ge a+b+ca2+b2+c2+43≥a+b+c