Solve a higher trigonometric math problem

By the Integrative Sum Difference Formula:

cosx cosy 1 2 cos (x y) cos (x y) and many more, you can search the Internet for your own and differential product or product and differential formula ......Solved cos (a b)cos(a b) 1 2 cos (2b) cos (2a) 1 2 1 2(sina) 2 1 2(sinb) 2 1 3

Simplify: (sin a) 2 (sin b) 2 2 3 Use the formula well, be familiar with the formula, and do more ......I feel like there's something to ......In the future, you can ask me about high school mathematics, but the most important thing is to test yourself, and I may only teach you the solution ...... some types of questionsMaster the ...... solutionIt's the same no matter how it changes.

a+2b=120°，a/2+ b=60°

So tan(a 2+ b) = 3, (tana 2+tanb) (1- tana 2 tanb) = 3, because tana 2 tanb = 2- 3 substitution: tana 2 + tanb = 3- 3

Simultaneous tana 2+tanb=3- 3 and tana 2 tanb=2- 3 can be obtained:

tana 2 = 1, tanb = 2-3 or tana 2 = 2- 3, tanb = 1.

So a = 2, b = 12 or a = 6, b = 4

A is an acute angle, so a = 2 and b = 12 are rounded.

a=π/6,b=π/4.

The direct solution yields -6<=a<=2

Because in the second quadrant, then a belongs to [2k +1 2 ,2k +

So -3 2 <=a<=- or 1 2 <=a<=2

y=sin 2(a)+sinb (a+b=90 degrees, sinb=sin(90°-a)=cosa)).

sin^2(a)+cos(a) (sin^2(a)＋cos^2(a)＝1)

cos 2(a) + cos(a) + 1 (let cosa be t, a (0,90), t (0,1)).

t^2+t+1

t-(1/2))^2+5/4

There is a maximum value when t belongs to (0,1), and when t=1 2, ymax=5 4

a+b=90 degrees sinb=cosa

y=sina squared + 2sinb = sina square + 2cosa = 1-cosa square + 2cosa = - (cosa-1) square + 2

0 degrees

tanα=y/x=0

Tan exists.

If the terminal edge falls on the y-axis, tan does not exist.

sinβ=2*tan(β/2)/[1+(tan(β/2)^2 ]cosβ=[1-(tan(b/2))^2]/[1+(tan(b/2))^2]

2) Since sin >0, cos >0, we know that belongs to (0, 2), and from the meaning of the title belongs to (0, ) so + belongs to (0, 3 2), but since sin( +=5 13>0, so + belongs to (0, ) If + belongs to (0, 2), then 2> +0, there should be sin( +=5 13 > sin =4 5, but this is not true, so + belongs to ( 2, so cos( +=- (1-(5 13) 2)=-12 13.

So sin = sin[(

sin(α+cosβ-cos(α+sinβ

There are only questions, no stems!

Because 1+sinx 1-sinx=(1+sinx) 2 (1-sinx)(1+sinx)=(1+sinx) 2 (1-sinx 2)=(1+sinx) 2 cosx 2

So cosx*root number(1+sinx 1-sinx)=cosx*(1+sinx) absolute cosx=1+sinx (positive cosx in quadrant 4).

Similarly, sinx * root number (1+cosx 1-cosx) = -1-cosx (sinx is negative in the fourth quadrant).

So f(x)=sinx-cosx=root2sin(x-4) and thus f(-4)=root2sin(-4-4)=-root22)x(2, )

then x-4 (4, 3, 4).

Then sin(x-4) (root2 2,1]then root2sin(x- 4) (1,root2]This is the answer.

f(x)=cosx*√[1+sinx)/(1-sinx)]+sinx*√[1+cosx)/(1-cosx)]

cosx*√[1+sinx)(1+sinx)/(1-sinx)(1+sinx)]+sinx*√[1+cosx)(1+cosx)/(1-cosx)(1+cosx)]

cosx * 1+sinx)/|cosx| +sinx*(1+cosx)/|sinx|

1)x (- 2,0), in the third quadrant, then cosx is negative and sinx is also negative.

Then f(x) = - 1-sinx-1-cosx = 2-sinx-cosx

f(-π/4)=2 -√2/2-√2/2=2-√2

2) x ( 2, ) is in the second quadrant, then cosx is negative and sinx is also positive.

Then f(x) = -1-sinx+1+cosx = cosx -sinx = - 2 sin(x- 4).

x ( 2, ) then x-4 ( 4, 3 4).

sin(x-π/4)∈(2/2,1)

Then - 2 sin(x- 4) (2, -1

That is, when x ( 2, ), the range of the function f(x) is (- 2, -1).

Let -3a = a 2a = b, then r = a +b = 9a +4a =13a r = root number 13a

sina = a r=-3 root number 13 cos=b r=2 root number 13 then sin 2 +cos *sin -1=9 13-6 13-1=3 13-1=-10 13