sinQcosQ is the square of x, 2xsina sinB, two roots of 0, and 2cos2a cos2B

It feels like there is a mistake in your problem, and the equation should be x 2-2xsina + sinb 2 = 0

quadratic equation ax 2 + bx + c = 0 followed by x1, x2; It has the following properties: x1+x2=-b a, x1*x2=c a

In this problem, sinqcosq is the square of x, 2xsina, sinb 2, and the two roots of 0.

So sinq+cosq=-2sina, sinq*cosq=sinb 2

Because sinq 2 + cosq 2 = 1

So sinq 2+cosq 2=(sinq+casq) 2-2*sinq*cosq=1

i.e. (-2sina) 2-2*sinb 2=1

4*sina^2-sinb^2=1

And because cos2a=cosa 2-sina 2=1-2sina 2 2sina 2=1-cos2a

So 2-2cos2a-2sinb 2=1

This question is proven.

There is no mistake in the question.

First tell a formula:

sin2q=2tanq (1+tan square q)=cos2q=(1-tan square q) (1+tan square q)=so sin square q + sinqcosq-2cos square q=-(cos square q-sin square q) + square q=-cos2q + square q-1)+

cos2q+==

Step by step, take a closer look......It is ...... general method

It is also possible to use tanq=2 to find sinq and cosq for direct generation.

sin square q + sinqcosq-2cos square q = (sin square q + sinqcosq-2cos square q) (sin square q + cos square q).

tan square q + tan q-2) (tan square q + 1), [the numerator and denominator are divided by cos square q).

4/5。Directly raise the cos squared out. After the proposal is the square multiplication of cos (tan squared plus two times tan minus two). Find out that the square of cos is 1 5. Just substitute it.

Because the two roots are equal, so.

36sin^2q-4tanq=0

9sin^2q-sinq/cosq=0

9sinq-1/cosq=0

9sinqcosq=1

There is sin 2q+cos 2q=1

sinq+cosq)^2=1+2sinqcosq=1+2/9=11/9

Because of 0, sinq+cosq=( 11) 3

x squared - 6xsinq + tanq = 0 (09sin q = sinq cosq

sinqcosq=1/9

sinq+cosq=root(sinq+cosq) = root(1+2sinqcosq).

Root number (1+2 9).

Two are equal i.e.

x²-6xsinq+tanq=0

x-3sinq)²=0=x²-6xsinq+9sin²q=x²-6xsinq+tanq

9sin²q=tanq

9sinq=1/cosq

9sinqcosq=1

sinqcosq=1/9

sinq+cosq)²=1+2sinqcosq=1+2/9=11/9

sinq+cosq= 11 3 under the root number

Since 00 cosq>0

So sinq + cosq = 11 3 under the root number

f(a)=1, 2 root number 3sinacosa+(cosa) 2-(sina) 2=1

2 roots: 3sinacosa+(cosa) 2-(sina) 2 roots: 3sin(2a)+cos(2a).

2sin(2a+π/6)

2sin(2a+ 6)=1, 2a+ 6=5 6a= 3 Counting this, I found that there is a problem with the question Zheng Tuanmu, if the envy is changed to f(c)=1, it is very reasonable.

(5x-4)(5x+3)=0 So x=4 5,x= -3 5 in(0,vulture),sin,cos is the root of the equation sin =4 5,cos 3 5 sin 2 -cos 2 =16 25-9 25=7 25 sin 3 -cos 3 =64 125+27 125=91 125 tan -cot =-4 3-3 4=-25 12

sinq*cosq

1/2[(sinq+cosq)^2-1]=1/2(1/4-1)=sinq-cosq)^2=

sinq+cosq)^2-4sinq*cosq1/4-4x(-3/8)=7/4

Because qe(0, root number 2), so when q is less than half of the pie, sinq small crack rent rushes to cosq, and q is greater than cosq when q is greater than half of pie

Hence sinq-cosq

7 2 or type.

1、a=

then a + b = 1

a+b=(√3+1)/2

ab=m, then (a+b) -2ab=1

4+2√3)/4-2m=1

m=√3/4

2. Original formula = sinq (1-cosq sinq) + cosq (1-sinq cosq).

sin²q/(sinq-cosq)+cos²q/(cosq-sinq)

sin²q-cos²q)/(sinq-cosq)=sinq+cosq

x²-(3+1)x+√3/2=0

So x=[( 3+1) (3-1)] 4x1=1 2, x2= 3 2

So q = 6 or 3

For convenience, we use x=sinq, y=cosq(1)(sinq) 2+(cosq) 2=1, so x 2+y 2=1

x-y=1/5

So according to (x-y) 2=x 2-2xy+y 2(1 5) 2=1-2xy

xy=12/25

Therefore, sinq·cosq=12 25

2) When 0y, y=4 5 and y=-3 5y=3 5 or -4 5 are discarded

So x=4 5 or 3 5

And 3 5-(-4 5) is not equal to 1 5, so y=-4 5 is rounded, so y can only make 3 5

So tanq=sinq, cosq=x, y=(4 5), (3 5)=4, 3