If SiNx + cosx = 1, find the value of 1-sin2x / cos? X-sin? X

If SiNx + cosx = 1, find the value of 1-sin2x / cos? X-sin? X

A:
sinx+cosx=1
Square the two sides to get:
sin²x+2sinxcosx+cos²x=1
2sinxcosx=0
sin2x=0
So: cos2x = - 1 or cos2x = 1
1-sin2x/(cos²x-sin²x)
=1-0
=1
(1-sin2x)/(cos²x-sin²x)
=(1-0)/(cos2x)
=1/cos2x
=-1 or 1

(1) If SiNx = 2cosx, then sin? X = (2) if Tan α = cos α is known, then sin α =? Ask brothers to write the detailed process Main question 1

The results show that: 1. SiNx = 2cosxcosx = SiNx / 2 substituting sin? X + cos? X = 1sin? X + (SiNx / 2) 2 = 15sin? 2x / 4 = 1sin? 2x = 4 / 52. Tan α = sin α / cos α = cos α sin α = cos? α sin? α + cos? α = 1sin? α + sin α = 1 (sin

If SiNx = 2cosx, then sin2x + 1 = 2___ .

∵sinx=2cosx，∴tanx=2．
Then sin2x + 1 = sin2x
sin2x+cos2x+1=tan2x
tan2x+1+1=22
22+1+1=9
5．
So the answer is: 9
5．

Y = cosx + SiNx = √ 2 (√ 2 / 2sinx + √ 2 / 2cosx) why = √ 2 (COS π / 4sinx + sin π / 4cosx)

Because cos π / 4 = √ 2 / 2, sin π / 4 = √ 2 / 2
That is to write √ 2 / 2 as cos π / 4 and sin π / 4 respectively

Simplification: (1) 1 / 2cosx - √ 3 / 2sinx (2) √ 3sinx + cosx (3) √ 2 (SiNx cosx) (4) √ 2cosx - √ 6sinx

（1）1/2cosx-√3/2sinx
=sinπ/6 cosx-cosπ/6 sinx
=sin(π/6-x)
（2）√3sinx+cosx
=2（√3/2sinx+1/2 cosx）
=2（cosπ/6sinx+sinπ/6cosx）
=2sin（x+π/6）
（3）√2(sinx-cosx)
=2（√2/2sinx-√2/2cosx）
=2（cosπ/4 sinx-sinπ/4 cosx）
=2sin(x-π/4)
（4）√2cosx-√6sinx
=2√2（1/2 cosx-√3/2 sinx）
=2√2（sinπ/6 cosx-cosπ/6 sinx）
=2√2 sin(π/6-x)
I hope it can help you! Wish you a happy study

Y = 1 / 2cosx + √ 3 / 2sinx 2. Y = SiNx cosx 3. Y = √ 3sinx + cosx

y=1/2cosx+√3/2sinx
=cos(x-π/3)
Maximum: 1
Minimum: - 1
y=sinx-cosx
=√2sin(x-π/4)
Maximum: √ 2
Minimum: - √ 2
y=√3sinx+cosx
=2sin(x+π/6)
2: maximum
Minimum: - 2

13. Simplification: (1) 3 √ 15sinx + 3 √ 5cosx (2) 3 / 2cosx - √ 3 / 2sinx (3) √ 3sinx / 2 + cosx / 2 （4）√2/4sin(π/4-x)+√6/4cos(π/4-x) （5）sin347°cos148°+sin77°cos58° （6）sin164°sin224°+sin254°sin314° （7）sin(a+b)cos(γ-b)-cos(b+a)sin(b-γ) （8）sin(a-b)cos(b-γ)-cos(a-b)sin(γ-b) （9）tan5π/4+tan5π/12 / 1-tan5π/12 （10）sin(a+b)-2sinacosb) / 2sinasinb+cos(a+b)

1）3√15sinx+3√5cosx
=6√5(√3/2sinx+1/2cosx )
=6√5sin(x+π/3)
2）3/2cosx-√3/2sinx
=√3(√3/2cosx-1/2sinx )
=√3cos(x+π/6)
3）√3sinx/2+cosx/2
=2(√3/2sinx/2+1/2cosx/2 )
=2sin(x/2+π/6)
4）√2/4sin(π/4-x)+√6/4cos(π/4-x)
=√2/2[1/2sin(π/4-x)+√3/2cos(π/4-x) ]
=√2/2sin(5π/12-x)
5）sin347°cos148°+sin77°cos58°
=sin13°cos32°+cos13°sin32°
=sin50°
6) Sin164 ° sin234 ° + sin254 ° sin314 °
=-sin16°sin54°+sin74°sin46°
=-sin16°sin54°+cos16°cos54°
=cos70°
7）sin(a+b)cos(γ-b)-cos(b+a)sin(b-γ)
＝sin(a+b)cos(γ-b)+cos(b+a)sin(γ-b)
=sin(a+b+γ-b)
=sin(a+γ)
8）sin(a-b)cos(b-γ)-cos(a-b)sin(γ-b)
=sin(a-b)cos(γ-b)-cos(a-b)sin(γ-b)
=sin(a-b-γ+b)
=sin(a-γ)
9）tan5π/4+tan5π/12 / 1-tan5π/12
=tanπ/4+tan5π/12 / 1-tan5π/12
=(tanπ/4+tan5π/12) / (1-tanπ/4*tan5π/12 )
=tan(π/4+5π/12)
=tan2π/3
=-√3
10）sin(a+b)-2sinacosb) / 2sinasinb+cos(a+b)
=[sinacosb+cosasinb-2sinacosb]/[cosacosb-sinasinb+ 2sinasinb]
=(cosasinb-sinacosb)/(cosacosb+sinasinb)
=sin(b-a)/cos(b-a)
=tan(b-a)

Let a = (2sinx, Radix 3sinx), B = (cosx, 2sinx), C = (2cosx, SiNx) (1) find a multiplication B and | b-c|

A * b = 2sinxcosx + (2 roots 3) sin ^ x
=Sin2x + Gen 3 - (Gen 3) cos2x
=2Sin (2x - π / 3) + radical 3
(|b-c|)^=b^+c^+2bc
=cos^x+4sin^x+4cos^x+sin^x+4cos^x+4sin^x
=9cos^x+9sin^x
=(9 radical 2) sin (x + π / 4)
|b-c| = radical [(9 radical 2) sin (x + π / 4)]

Known vector a = (1-cosx, 2sinx / 2), B = (1 + cosx, 2cosx / 2) (1) If f (x) = 2 + sinx-1 / 4 | A-B ^ 2, find the expression of F (x) (2) If the images of function f (x) and function g (x) are symmetric about the origin, find the analytic formula of function g (x) (3) If h (x) = g (x) - YF (x) + 1 is an increasing function on [- π / 2, π / 2], find the value range of real number y

1)a-b=(-2cosx,2sinx/2-2cosx/2)
f(x)=2+sinx-(1/4)[4cos²x+4(sin²x/2+cos²x/2-2sinx/2cosx/2)]
=2+sinx-cos²x-(1-sinx)
=2+sinx-(1-sin²x)-1+sinx
=sin²x+2sinx
2) Let the point (x, y) on G (x), then the point corresponding to f (x) is (- x, - y)
∴-y=sin²(-x)+2sin(-x)=sin²x-2sinx
∴y=-sin²x+2sinx
That is g (x) = - sin? X + 2sinx
The next λ is y
3)h(x)=(-sin²x+2sinx)-λ(sin²x+2sinx)=(-1-λ)sin²x+(2-2λ)sinx
T = SiNx increases monotonically at [- π / 2, π / 2],  H (x) increases monotonically on (- 1 - λ) t  2 + (2-2 λ) t on [- 1,1]
H (x) is a quadratic function about t, and the axis of symmetry is t = (2-2 λ) / 2 (1 + λ) = (1 - λ) / (1 + λ)
If - 1 - λ > 0, λ

Given the vector M = (Radix 3sinx + cosx, 1), n = (f (x), cosx), and M / / n (1) Find the monotone interval of function f (x); (2) Let a, B and C be the opposite sides of a, B and C of the inner angles of the triangle ABC. If f (A / 2) = 1 / 2 + root 3 / 2, a = 1, B = root 2, calculate the area of triangle ABC I look forward to your answers to my other questions, which are those with 30 points.

(1) Therefore (√ 3sinx + cosx) cosx x-f (x) = 0, that is, f (x) = 0, that is, f (x) = (3 / 2) sin2x + (1 / 2) cos2x + 1 / 2 = sin (2x + π / 6) + 1 / 2 makes - π / 2 + 2K π ≤ 2x + π / 6 ≤ π / 2 + 2K π solution - π / 3 + K π ≤ x ≤ π / 6 + 6 + K π, namely, the increasing interval is [- π / 3 + 3 + K π ≤ x ≤ π / 6 + 6 + K π, that is, the increasing interval is [- π / 3 + 3 + k k k k k k k k k + π - π - π / 3 + K K K + K π