On the problem of trigonometric function, we know that 3sin ^ 2A + 2Sin ^ B = 2sina, find the value range of sin ^ 2A + sin ^ B

On the problem of trigonometric function, we know that 3sin ^ 2A + 2Sin ^ B = 2sina, find the value range of sin ^ 2A + sin ^ B

Sin ^ 2A + sin ^ B = Sina sin (2a) / 2 = Sina sinacosa, if f (a) = Sina Sina sinacosa, the derivative of F '(a) = cosa - ((COSA) ^ 2 - (Sina) = - 2 (COSA) ^ 2 + cosa + 1 = (2cosa + 1) (1-cosa) when a belongs to (0,2 / 3 π) or (4 / 3 π, 2 π), f (a) is an increasing function and a belongs to (2 / 3 π, 4 /...)

It is known that ∠ A is an acute angle, Tana COTA = 4, Tan ^ 2A + cot ^ 2A =?

Half angle formula: Tana = (1-cos2a) / sin2acota = (1 + cos2a) / sin2atana COTA = - 2cos2a / sin2a = - 2cot2a = 4, then cot2a = - 2 obviously tan2a = - 1 / 2, so the answer is - 5 / 2, which is twice. If you want to square the two sides of the known formula, because Tana * COTA = 1, the result is equal to the square of 4

On sin, cos, Tan, cot, CSC, Sec coty-tany ----------=The quadratic power of CSC y-sec sinycosy

(coty-tany)/(sinycosy)
=(cosy/siny-siny/cosy)/sinycosy
=(cos^2y-sin^2y)/sin^2yos^2y
=1/sin^2y-1/cos^2y
=csc^2y-sec^2y

{[xˆ(-2)+yˆ(-2)]/[xˆ(-2/3)+yˆ(-2/3)]}-{[xˆ(-2)-yˆ(-2)]/xˆ(-2/3)-yˆ(-2/3)}

The original formula = [x ^ (- 2 / 3) + y ^ (- 2 / 3)] [(x ^ (- 4 / 3) - x ^ (- 2 / 3) y ^ (- 2 / 3) + y ^ (- 4 / 3)] / (x ^ (- 2 / 3) + y ^ (- 2 / 3) - [x ^ (- 2 / 3) - y ^ (- 2 / 3)] [(x ^ (- 4 / 3) + x ^ (- 2 / 3) y ^ (- 2 / 3)] / [x ^ (- 2 / 3) - y ^ (- 2 / 3)] = x ^ (-...)

sin^2xtanx+(cos^2x/tanx)+2sinxcosx-1+cos/sinxcosx

cos/sinxcosx ?

2 sin? X-cos? X + sin xcosx-6sinx + 3cosx = 0 is decomposed into (2sinx cosx) (SiNx + cosx-3) = 0 How do you get this? How do you decompose this formula in the future?

2Sin? X-cos? X + sinxcosx-6sinx + 3cosx = (2Sin? X-sinxcosx) + (2sinxcosx-cos? X) - (6sinx-3cosx) = SiNx (2sinx cosx) + cosx (2sinx cosx) - 3 (2sinx cosx) = (2sinx cosx) (SiNx + cosx-3) = 0

(2 / 3 of 2A * 1 / 2 of B) (- 6a to 1 / 2 * b) / (- 3a to the 6th * B to the 5th) Look at the word is very much, actually very simple

tupian

Simplification: 1 - [sin6 α + cos6 α] "6" is the sixth power The solution is as follows: the square of the original formula = 1 - {[sin2 α + Cos2 α] - 3 sin2 α Cos2 α} [sin2 α + Cos2 α] "2" is quadratic It's explained in the book. I can't understand it~

1-(sin^6a+cos^6a)
According to a ^ 3 + B ^ 3 = (a + b) (a ^ 2-A * B + B ^ 2)
=1-[(sin^2a+cos^2a)*(sin^4a-sin^2acos^2a+cos^4a)]
According to sin ^ 2A + cos ^ 2A = 1
=1-sin^4a-sin^2acos^2a-cos^4a
=（sin^2a+cos^2a）-sin^4a+sin^2acos^2a-cos^4a
=（sin^2a-sin^4a）+（cos^2a-cos^4a）+sin^2acos^2a
Because sin ^ 2a-sin ^ 4A = sin ^ 2A (1-sin ^ 2a)
=sin^2a*cos^2a
Similarly, cos ^ 2A cos ^ 4A = sin ^ 2A * cos ^ 2A
Because 2sina * cosa = sin2a
So the original formula = 3sin ^ 2A * cos ^ 2A

Calculation 1 sin50°+ Three cos50°=______ .

1sin50°+3cos50°=cos50°+3sin50°sin50°cos50°=2(12cos50°+ 32sin50°)sin50°cos50°=2(sin30°cos50°+  cos30°sin50°)sin50°cos50°=2sin(30°+50°)sin50°cos50°=2sin80°sin50°cos5...

How to simplify cot (3 π + α)? The trouble process is clear!

First of all, cot is a function with a period of PI, which you can see in the graph
So every time you subtract PI or add PI, the value doesn't change
therefore
cot（3π+α）=
cot（2π+α）=
cot（π+α）=
cot（α）
The other is to use the sum angle formula, but because it is an integral multiple of pi / 2, it is not necessary~
If you don't understand, keep asking~