It is known that Tana = - 4 divided by 3, and a is the fourth quadrant angle cosa = Sina=

It is known that Tana = - 4 divided by 3, and a is the fourth quadrant angle cosa = Sina=

sina=-3／5 cosa=4／5
If the point P (Tana, COSA) is known to be in the third quadrant, then which quadrant is the final edge of angle a
Analysis:
Because point P (Tana, COSA) is in the third quadrant
So Tana
What are the maximum and minimum values of y = SiNx + cosx on [- 90 °, 90 °]
y=sinx+cosx
=√2sin(x+45°)
Maximum = √ 2
Minimum = - 1
The cube of [b (- cube of a-cube of a-cube of a)]
Urgent! Ask the great gods to speed up the solution, there should be a process, thank you, OK, will add points!
The cube of [b (- cube of a-cube of a-cube of a)]
＝[b（-3a^3)^2]^3
=[b*9a^6]^3
=729a^18b^3
The function f (x) = (cosx) 3 (SiNx) 2-cosx has the maximum value of [0,2 π]
Function f (x) = cosx ^ 3 + SiNx ^ 2-cosx,
Mr. / MS's answer is incorrect, and his / her analysis needs careful consideration,
"When x = π / 2, the former is the maximum, and the latter is also the maximum". How can the latter 1-cos X be the maximum? Did you not expect that cos x would be negative?
Verify with x = 1.8, it is easy to find that f (x) > 1
Let cos x = t, then the original problem is converted to - 1
The square of (a-b) · (a + b) · (a-b) · (B-A) · (a + b)
Shanghai homework grade 7 Volume 1 page 33 question 8
There is also this question: &# 189; m · Mn · M-M · n · M & # 178; + &# 189; M & # 178; · n · M
Solution (a-b) (a + b) (a-b) &# 178; (B-A) &# 179; (a + b) &# 178; = - (a + b) &# 179; (a-b) (a-b) &# 178; (a-b) &# 179; = - (a + b) &# 179; (a-b) ^ 61 / 2m × Mn × M-M × n × m ^ 2 + 1 / 2m ^ 2 × n × M = 1 / 2nm ^ 3 + 1 / 2nm ^ 3 = (1 / 2-1 + 1 / 2) nm ^ 3 = 0 × nm ^ 3 = 0
The minimum value of [sincoy + 2] and the maximum value of [sincoy + 2] / X ∈
y=sinx+cosx
=√2sin(x+π/4)
x∈【0,π/2】
x+π/4∈【π/4,3π/4】
When x + π / 4 = π / 4 or 3 π / 4, the minimum value of Y is 1
When x + π / 4 = π / 2, the maximum value of Y is √ 2
y=sinx+cosx=√2sin（x+π/4)
x+π/4∈[π/4,3π/4]
Then sin (x + π / 4) ∈ [√ 2 / 2,1]
So ymax = √ 2
ymin=1
y=sinx+cosx
=√2(√2/2*sinx+√2/2*cosx)
=√2(sinxcosπ/4+cosxsinπ/4)
=√2sin(x+π/4)
x∈[0，π/2]
x+π/4∈[π/4，3π/4]
When x = π / 4, sin (x + π / 4) = 1
ymax=√2
When x = π / 2, sin (x + π / 4) = √ 2 / 2
ymin=√2/2*√2=1
In real numbers 3.14, 3.3333l, 0, 0.412, 12 cycles, 0.10111011110 There are () irrational numbers
A2, B3, C4, D5
You should choose A2
The maximum value of function f (x) = (cosx) ^ 3 + (SiNx) ^ 2-cosx on [0,2 π] is?
If f (x) = cosx ^ 3 + SiNx ^ 2-cosx = cosx ^ 3-cos ^ 2-cosx + 1, let t = cosx, then f (T) = T ^ 3-T ^ 2-T + 1 (t ∈ [- 1,1]) f '(T) = 3T ^ 2-2t-1, Let f' (T) = 0 - > t = - 1 / 3 or 1 [- 1, - 1 / 3) (- 1 / 3,1] + - increase and decrease, so f (T) takes the maximum value at t = - 1 / 3, which is also the maximum value f (- 1 / 3) = 32 / 27
In the following real numbers, the irrational number is ()
³ √ 9, - 1 / 7, π / 2, - 3.14, √ 0.1,12, ³ √ - 27,0,0.3232232223 (a 2 is added between two adjacent 3 in turn),
³ √ 9,2 π, √ 0.1,0,0.3232232223 (a 2 is added between two adjacent 3 in turn),