To know 1 cos sin sin 0, 1 cos cos sin cos 0, then sin

1+cos(a)-sin(b)+sin(a)*sin(b)=0

1-cos(a)-cos(b)+sin(a)*cos(b)=0

Available. 1+cos(a)=(1-sin(a))*sin(b)

1-cos(a)=(1-sin(a))*cos(b)

Square on both sides. 1+2*cos(a)+cos(a)^2=(1-sin(a))^2*sin(b)^2

1-2*cos(a)+cos(a)^2=(1-sin(a))^2*cos(b)^2

Add the two equations together.

2+2*cos(a)^2=(1-sin(a))^2

Sorted out. 3*sin(a)^2-2*sin(a)-3=0

Solve the equation to get.

sin(a) = 1 3-1 3*(10) (1 2) or 1 3+1 3*(10) (1 2).

Since -1<=sin(a)<=1 (<= is greater than or equal to, the commonly used symbol cannot be played).

So sin(a)=1 3-1 3*(10) (1 2).

Test: Bring sin(a) back to the original equation and the two equations hold.

Subtracting formula 2 from Eq. 1 gives sin = cos, so sin = 1 2

1+sin2 =-3cos2, so sin2 +cos2 +2sin cos =3cos2 -3sin2, spring stove closed 4sin 2 -2cos 2 +2sin cos =0, so 4tan 2 -2+2tan =0, split 0, 2 ) tan =2;Therefore, the reply to the case is: 2

Because: aob=1 4 aod, aoc=1 2 aod, and boc=15°

So the bundle is combined: ob bisects the AOC

So: aob = boc = 15°

So: AOC=2*15°=30°

The object is covered with Zheng Chang: aod=2 aoc=60°

2sin -cos =1, (2sin -cos) =1, 4sin -4sin cos +cos =1, and because sin +cos = 1, so 3sin -4sin cos = 0, so 3sin = 4cos, sin = 4 3cos, bring in the 2sin -cos = 1 equation, solve the equation to get cos = 3 7, sin = 4 7, sin +cos +1 = 2, sin -cos +1 = 8 7, (sin +cos +1(sin -cos +1)=7 4,I'm a mouth counter.,Landlord, take a look first!

Math isn't very good, I don't know if it's right or not.

From 2sin -cos =1, sin -cos = 1-sin can be deduced, 2sin =1+cos, thus simplifying (sin +cos +1) (sin -cos +1)=3sin 2+sin. It is also thought that sin square + cos square = 1, and sin = 4 5 or 0 can be obtained. Substituting the above equation can be used to obtain the result.

Equation 1 is sin +cos = (root number 3-1) Formula 22 is the square of sin + the square of cos is equal to 1 solution sinx = root number 3 2, cosx = -1 2 then tan = is - root number 3

Answer: 2 of the known questions should be

tan²α-2tan²β-1=0

tan²α+1=2(tan²β+1)

sin²α/cos²α+1=2(sin²β/cos²β+1)∴ sin²α+cos²α)/cos²α=2(sin²β+cos²β)/cos²β

1/cos²α=2/cos²β

cos²β=2cos²α

2cos²β=2*2cos²α

1+cos2β=2(1+cos2α)

i.e. cos2 =2cos2 +1

Please write clearly on the square: (1+2sin cos ) (cos -sin )= (1+tan ) (1-tan )

The left numerator = cos +sin +2sin cos = (cos +sin).

Left denominator = cos -sin = (cos +sin) (cos -sin)

So the left end = (cos +sin ) (cos -sin ) = (1+tan ) (1-tan ) = right end.

cos2θ=cos²θ-sin²θ=cos²θ-1-cos²θ)=2cos²θ-1

Proof: Equation right = (1+tan ) (1-tan ) = (cos + sin ) (cos -sin ) = (cos + sin ) (cos -sin ) (cos + sin ) (cos + sin )

cosθ+sinθ)²/(cos²θ-sin²θ)

1+2sinθcosθ)/cos2θ

Then substitute according to any value (e.g. =30°) to obtain.

The title is incorrect. By the way, please write the square clearly.

The title is (1+2sin cos) (cos -sin ) is right!

Proof: Left side of the equation = (1+2sin cos) (cos -sin) = (cos + sin) (cos -sin).

cosθ+sinθ)²/(cosθ-sinθ)(cosθ+sinθ)

cosθ+sinθ)/(cosθ-sinθ)

cosθ≠0

The numerator and denominator are divided by cos.

Original = (1+tan) (1-tan) = right side of the equation.

The original formula is proven.

Answer: Because: 1 3sin -cos = 0, so :tan = 3

So: (2cos2 +1) 2sin2 = [2(cos) 2-2(sin) 2+(sni) 2+(cos) 2] (4sin cos).

3(cosθ)^2-(sinθ)^2]/[4sinθcosθ]=[3-(tanθ)^2]/[4tanθ]=(3-9)/12=-1/2

acosθ-sinθ)²a²cos²θ+sin²θ-2asinθcosθ=1……①

asinθ+cosθ)²a²sin²θ+cos²θ-2asinθcosθ=1……②

Obtain: a cos +sin +a sin +cos = 2, i.e. a (sin +cos ) sin +cos )2a +1 = 2

Solution: a = 1

1° If a=1, then cos -sin =1......③sinθ+cosθ=1……④

Get: 2sin = 0

Answer: sin = 0

2° if a = -1

then -cos -sin = 1......⑤

sinθ+cosθ=1……⑥

Get: -2sin = 2

Solution: sin = -1

Middle School Students' Mathematics, Physics and Chemistry] team wdxf4444 is your answer! Good luck with your learning.

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