If cos (π / 2-A) = 3 / 5, a ∈ (π / 2, π), Tana=

If cos (π / 2-A) = 3 / 5, a ∈ (π / 2, π), Tana=

cos(π/2-a)=sina=3/5
sin²+cos²a=1
A ∈ (π / 2, π), then cosa
-3/4
cos(pi/2-a)=3/5
So sin (a) = 3 / 5
Then cos (a) = 4 / 5
So tan (a) = 3 / 4
Cosa = - 4 / 5, a in the third quadrant, find (1 + Tana / 2) / (1-tana / 2)
sina
Cosa = - 4 / 5, a in the third quadrant Sina = - 3 / 5
（1+tana/2)/(1-tana/2)
=(1+sina/2/cosa/2)/(1-sina/2/cosa/2)
=(cosa/2+sina/2)/(cosa/2-sina/2)
=(cosa/2+sina/2)^2/(cosa/2-sina/2)(cosa/2+sina/2)
=(1 + Sina)... Expansion
Cosa = - 4 / 5, a in the third quadrant Sina = - 3 / 5
（1+tana/2)/(1-tana/2)
=(1+sina/2/cosa/2)/(1-sina/2/cosa/2)
=(cosa/2+sina/2)/(cosa/2-sina/2)
=(cosa/2+sina/2)^2/(cosa/2-sina/2)(cosa/2+sina/2)
=(1+sina)/cosa
=(1-3/5)/(-4/5)
=-1 / 2 question: (COSA / 2-sina / 2) (COSA / 2 + Sina / 2) = what are the specific steps of cosa? What formula is used?
Given 3 times the square of X - x = 1, find 6 times the cube of X + 7 times the square of X - 5 times x + 2007
6x^3+7x^2-5x+2007 =2x(3x^2-x)+9x^2-5x+2007 =2x*1+9x^2-5x+2007 =9x^2-3x+2007 =3(3x^2-x)+2007 =3*1+2007 =2010
Given the function f (x) = 2Sin (Wx + π / 6) (w ＞ 0), the distance between the image of y = f (x) and the two adjacent intersections of the line y = 2 is equal to π, then the monotone increasing region of F (x)
F (x) = 2Sin (Wx + π / 6) ∵ the distance between the image of the function and the two adjacent intersections of the line y = 2 is equal to π ∵ the period of the function T = π = 2 π / w ∵ w = 2 ∵ f (x) = 2Sin (2x + π / 6) such that - π / 2 + 2K π ≤ 2x + π / 6 ≤ π / 2 + 2K π K ∈ Z, the monotonic increase of K π - π / 3 ≤ x ≤ K π + π / 6 K ∈ ZF (x) is obtained
Square + (- 4AB & # 179;) × (- AB)
(3ab^2)^2+(-4ab^3)*(-ab)
=9a^2b^4+4a^2b^4
=13a^2b^4
The number of removable discontinuities of function f (x) = (x-x & # 178;) / sin π x ()
Two X = 0 and x = 1
If the square of a + 3AB = - 4, the square of B + AB = 6, then the square of a + 4AB + the square of B=
Square of a + 4AB + square of B
=Square of a + 3AB + square of B + ab
=-4+6
=2
Ten
solution
a^2+4ab+b^2
=a^2+3ab+b^2+ab
=-4+6
=2
-ab
Finding the period of function: F (x) = Sin & # 178; X
f(x)=sin²x=(-1/2)(1-2sin²x)+1/2=-(1/2)cos2x+1/2
So f (x) period is: π
What is the sum of [(- 1) cube times the square of a] 4?
The eighth power of a
It is known that SiNx cosx = (√ 2) / 3 (0
(sinX-cosX)^2=2/9
2sinXcosX=7/9
(sinX+cosX)^2=1+2sinXcosX=16/9
SiNx + cosx = 4 / 3 or = - 4 / 3
Because 0
Because 00
SiNx cosx = (√ 2) / 3 square
sin^2 x -2sinxcosx + cos^2 x =2/9
1-2sinxcosx=2/9
2sinxcosx=7/9
1+2sinxcosx=1+7/9=16/9
sin^2 x+cos^2 x+2sinxcosx=16/9
(sinx+cosx)^2=16/9
sinx+cosx=4/3