A few trigonometric questions!! Please, everyone, come on

sin(110°+a)=sin(90°+20°+a)=cos(20°+a)=-3 5, and 70° a 160°, then sin(20°+a) = 1-(-3 5) 2=4 5

sin(x+y)=1 know, x+y= 2+2k, then 2x+2y= +4k, 2x+3y= +4k +y, tan(2x+3y)=tan[( 4k )+y)]=[tan( +4k )+tany] [1-tan( +4k )tany], and tan( +4k )=0, substitute it into the above equation to get the original formula =tany

This question should be a bit of a problem for you: the sin item disappears after the formula to be simplified, and you check it again.

But I still calculated a few steps, and it should be useful: tan = (a 1-a), then tan 2 = a 1-a, 1+tan 2 =1+a 1-a=1 1-a=sec 2 =1 cos 2, then cos 2 = 1-a, substitute it, and you look at ...... again

1.Think of 20°+a as (110°+a)—90°2().

2.Since sin(x+y)=1 then cos(x+y)=0, then sin2(x+y)=0, cos2(x+y)=1, tan2(x+y)=0, tan2x+3y=2(x+y)+y, then use the formula. ok

3.The formula is somewhat incomprehensible...

sin +cos = root number m

sin *cos =1 search clear only m

sinθ+cosθ)^2-2sinθ*cosθ=1m-2/m=1

m^2-m-2)/m=0

So Sepei m 2-m-2=0

m-2)(m+1)=0

So m=2 m=-1 (exclude).

The posture is sin =cos = 2 2

So =4+k

f(x)=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6)

t=2π/2=π

2kπ-π2≤2x+π/6≤2kπ+π2

2kπ-2π/3≤2x≤2kπ+π3

k - 3 x repentance seep k + 6

Increasing interval [k - 3, k + 6].

k is the number of bends of the whole blue teasing spine.

(cosa-sina) 2=1-2sina*cosa=1-2*1 8=3 4 cosa-sina = plus or minus 2 root number 3 And because 0a45 cosa-sina is greater than 0 cosa-sina = 2 points of the root number 3

I'm also anxious, where is the topic!

By analytic formula a(file stuffy 1,0) b(0,1).

So AOB is an isosceles right triangle, so P is the midpoint of AB.

Because oa=ob=1, ab=root2

The stupid friend is named ap=root number 2 2

Because OA=1

So op = root number 2 and Zhenghuai 2

So cos = op oa = root number 2 2

Let the perpendicular line through point b intersect ac at point o, then abc = abo- cbo

Because tanabo=1 tancbo=1 3

So tanabc=tan(abo=cbo)=(tanabo-tancbo) (1+tanabo*tancbo)=1 2

Let abc= , and the small corner on the right is ;

Then: =( 4)-

tanα=tan(2α/2)

1-cos2) sin2 (half-angle formula) = [1-cos( 2-2 )] sin( 2-2 ) = (1-sin2) cos2

1-2sin cos) [(cos) 2-(sin) 2] (doubling formula).

where sin = 10 10 and cos = 3 10 10 substitute it to obtain:

tan = (1-2 3 10) (9 10-1 10) = 1 2, so the tangent of abc is 1 2

Establish. A is an acute angle, sina<1, sina-1<0, and the absolute value of the square should be taken as the opposite number.

Under the root number (sin -1) is definitely a non-negative number, and then there is it anyway: sin <1 (because of the acute angle, it can't be taken at 90°).

So it can be directly equal.

It's founded! Acute angles are not true. Because sin has to take the range [-1,1].

Establish. Under the root number (sin -1) = sin -1 at absolute value, known to be an acute angle, then sin 1, then sin -1 0, then sin -1 = 1-sin at absolute value

So under the root number (sin -1) = 1-sin

Established because sina is less than 1.

(1)f(x)=2sin²(wx+π/4)+2cos²wx (w>1)

1-cos(2wx+π/2)+1+cos(2wx)

2+sin(2wx)+ cos(2wx)

2+√2sin(2wx+π/4)

t=2π/|2w|=2π/3

w=±3/2,w>1,w=3/2,f(x)= 2+√2sin(3x+π/4)

When sin(3x+ 4)=1, f(x) has a maximum value f(x)max=2+ 2, then 3x+ 4=2k + 2, and x=(2k + 4) 3=2k 3+ 12, k n;

2) The image of y=f(x) is shifted to the right 8 unit lengths, y=f(x- 8) = 2+ 2sin[3(x- 8)+ 4]=2+ 2sin(3x- 8), after symmetry along the y-axis, g(x)=2+ 2sin(-3x- 8)=2- 2sin(3x+ 8), let t= sin(3x+ 8), 2- 2t is the subtraction function, the monotonicity of the composite function can be obtained, 2k - 2 3x+ 8 2k + 2, t is the increase function about x, g(x)

is about the subtraction function of x, and the monotonic reduction interval of g(x) is [2k 3-5 24, 2k 3+ 8], k n