Senior 2 Mathematics Trigonometry, Senior 2 Trigonometry

1+2sinacosa=(sina + cosa) 2 because 1 sin is squared + cos is squared. So that's the perfect square formula.

So the molecule (1+sina + cosa + 2sinacosa) = (sina + cosa) (1 + sina + cosa).

Then the latter one goes about going with the denominator, which is sina + cosa

Denominator: 1+2sinacosa+sina+cosa=(sina) 2+(cosa) 2+2sinacosa+sina+cosa

sina+cosa)^2+sina+cosa=(sina+cosa)(1+sina+cosa)

And the molecule is 1+sina + cosa

The result of the upper and lower approximate differentiation was sina + cosa

The proposition is proven. Hope.

By the isoangular formula sina*sina + cosa*cosa = 1 molecule. 1+sina+cosa+2sinacosa=sina*sina+cosa*cosa+sina+cosa+2sinacosa=(sina+cosa)^2+sina+cosa=(sina+cosa)*(sina+cosa+1)

Divide by the denominator 1 + sina + cosa

The final result is sina + cosa, and it is proven.

Then you solve the problem and divide both sides of the equation by c, and you can get that the cosine value of c is negative, so it is an obtuse angle.

By y=cos2(x+4)-sin2(x+4)y=cos(2x+2).

y=-sin2x

Therefore, this function is an odd function that is the smallest of macro-envy and is pretending to be indefinite, so A

Choose ae This is a formula, a formula, a formula without deformation.

cos2a=cos^a-sin^a

Hints at this point, you'll do it in a hidden grip, right??

Strange and suspicious couple··· Because 2 is added, it becomes sin, and it is -sin, so it is an odd function and can be guessed.

By the title, f(x)=2(sinxcos 6+cosxsin 6)-2cosx

3sinx+cosx-2cosx

3sinx-cosx x∈[π/2,π]1. ∵x∈[π/2,π]

If sinx=4 5, then cos=-3 5

f(x)=(4√3-3)/5

2. f(x)=√3sinx-cosx

2sin(x-π/6)

x∈[π/2，π]

x-π/6)∈[/3,5π/6]

sin(x-π/6)∈[1/2,1]

The range of f(x) is [1,2].

(1) f(x)=(root number 3)*sinx-cosx, and x belongs to [pi 2,pi], so cosx<0, since sinx=4 5, so cosx=-3 5, substitution, f(x)=4 (root number 3) 5+3 5;

2) From f(x)=(root number 3)*sinx-cosx, it can be reduced to f(x)=2sin(x-pi 6), from x belongs to [pi 2,pi], and (x-pi 6) belongs to [pi 3,5pi 6], so we can know that the f(x) value range is [1,2].

According to my experience in doing problems, when encountering inconsistencies in variables in trigonometric functions, it is generally necessary.

1:f(x)=2(sinxcos30°+cosxsin30°）-2cosx

Root 3sinx-cosx

x is between [pi 2, pi].

sinx=4 5, so cosx=-3 5

f(x)=/5

2: f(x)=root3sinx-cosx=2sin(x-30°) in the range [1 2,1].

Suppose the radius is a

Then the arc length is 4-2a

The radian is 4 a-2

Area = a 2 * (4 a - 2) blind chain 2 = 2a - a 2 = 1a 2-2a + 1 = 0

a=1 The radian of the central angle of the reed is 2, and the chord length is 2*sin(1)=

Solution: Obtained by cos( sin(

cos cos sin sin sin cos sun sec cos sin

i.e. cos cos cos sin sin cos sin sin

cosα(cosβ＋sinβ)＝sinα(cosβ＋sinβ)…Because both are acute angles, cos is sin ≠0 and cos sin

Hence tan 1

If you're not in a hurry, I can show you tomorrow.