Given that a is the second quadrant angle, Sina = 3 / 5, what is tan2a equal to?

Given that a is the second quadrant angle, Sina = 3 / 5, what is tan2a equal to?

∵ A is the second quadrant angle, Sina = 3 / 5
Using unit circle to construct right triangle, Tana = - 3 / 4 can be obtained
∴tan2A=2tanA/(1-tan²A)=－24/7
In the triangle ABC, the vector AB = a, AC = B, and AE = 2 / 1Ac, try a, B to represent the vector BC
BC = ac-ab = b-a. there's something wrong with your input, but point E is useless
It is proved that if Liman = a, limbn = B, then LIM (an * BN) = a * B
This is true. This is the limit algorithm
Square of root 3 - square of root 2
Square of root 3 - square of root 2
=3-4
=-1
-1
Negative one
V 1 and V 2 are vector spaces over real number fields. It is proved that v 1 and V 2 are also vector spaces over real number fields
Let a and B belong to V1 intersection V2, K and l be any real numbers, then Ka + LB belongs to V1 intersection V2, so V1 intersection V2 is also a vector space over real number field
The minimum positive period of F (x) = root sign (COS & # 178; the fourth power X of x-cos)
(COS & # 178; the fourth power of x-cos)
=cos²x（1-cos²x）
=cos²xsin²x
f(x)=+-cosxsinx
=+-1/2sin2x
The minimum positive period is Pai
Liman = a, limbn = B is LIM (an + BN) = a + B what condition?
Is the necessary and sufficient conditions for those types of answers
Can you give me a reason,
Please look at the picture:
Square x minus 6x is eleven
x²+6x＝11
x²+2×3 x+3²＝11+3²
x²+3²=20
x+3 =±√20
x+3=2√5.x+3=－2√5
X1=2√5+3 X2＝3-2√5
Linear algebra problem: why is the necessary and sufficient condition of injectivity F-1 (02) = 01, where F-1 (02) represents the kernel of F (kerf), and 01,02 is the set V1, V2
If f is a injective from V1 to V2, we can see from the definition of injectivity that the images of different elements in V1 must be different. In V1, except 01, other elements are not equal to 0, so F-1 (02) = {01}. Conversely, to prove that f is a injective from V1 to V2, that is to say, the images of different elements in V1 must be different
Given the function f (x) = 1 / 2cos ^ x + root sign 3 / 2sinxcosx-3 / 4. Find the monotone decreasing interval of F (x)
Urgent age urgent
y=1/2cos^x+√3/2sinxcosx+1=(1+cos2x)/4+√3/4sin2x+1
=(√3/4sin2x+1/4cos2x)+5/4
=1/2sin(2x+π/6)+5/4
So 2K π - π / 2
See above.