1-2sin (π - 3) cos (π + 3)

1-2sin (π - 3) cos (π + 3)

The original formula = 1 + 2Sin (3) cos (3)
=1+sin6
Odd change and even invariance, symbols look at quadrant
　sin（π+α）= -sinα 　　cos（π+α）= -cosα 　　tan（π+α）= tanα 　　cot（π+α）= cotα

Simplification 1−2sin(π+3)cos(π+3)=______ .

1−2sin(π+3)cos(π+3)
=
1−2sin3cos3
=
sin23−2sin3cos3+cos23
=
(sin3−cos3)2
∵sin3＞cos3
Qi
(sin3−cos3)2
=sin3-cos3．
So the answer is: sin3-cos3

It is known that Tan α = - 1 3，cosβ＝ Five 5, α, β ∈ (0, π), then α + β=______ .

0

If Tan α = radical 2 / 2, find 2Sin α + cos α / 2Sin α - cos α

tana=√ 2/2
(2sina+cosa)/(2sina-cosa)
=(2sina/cosa+1)/(2sina/cosa-1)
=(2tana+1)/(2tana-1)
=(2*√ 2/2+1)/(2*√ 2/2-1)
=(√ 2+1)/(√ 2-1)
=3+2√ 2.

It is known that sin θ + cos = the root of 3 is 20

sinθ + cosθ = √2/3sinθ= √2/3 - cosθ sin^2θ= 2/9 - 2√2/3 cosθ + cos^2θ1-cos^2θ= 2/9 - 2√2/3 cosθ + cos^2θ2cos^2θ - 2√2/3 cosθ - 7/9 = 0(√2cosθ+1/3)(√2cosθ-7/3) = 00

Given sin θ + cos θ = negative fifth root 10 (0 < α < π), find the value of 1. Tan θ (2) 1 / sin θ + 1 / cos θ

Sin θ + cos θ = negative fifth root 10
Square gain
(sinθ+cosθ)²=2/5
sin²a+2sinacosa+cos²a=2/5
1+2sinacosa=2/5
sinacosa=-3/10
sinacosa/(sin²a+cos²a)=3/10
Divide the numerator and denominator by cos? A at the same time
tana/(1+tan²a)=3/10
3tan²a+10tana+3=0
(3tana+1)(tana+3)=0
Sin θ + cos θ = the root of the negative fifth 10 < 0
3π/4<α<π
tana=-1/3
1/sinθ+1/cosθ
=(sina+cosa)/sinacosa
=(- radical (10) / 5) / (- 3 / 10)
=2 roots (10) / 3

Sin (α + π / 3) + sin α = negative 5 / 4 radical 3 α ∈ (- π / 2,0) to find cos α How to change 9 / 4 * sin? 2 α = 9 / 4 (1-cos? α) = 3 / 4 * cos? A + 12 / 5 * cosx + 48 / 25 to cos? α + 4 / 5 * cos α - 11 / 100 = 0 sin(α+π/3)+sinα=-4√3/5 sinαcosπ/3+cosαsinπ/3+sinα=-4√3/5 3/2*sinα+√3/2*cosα=-4√3/5 3/2*sinα=-√3/2*cosα-4√3/5 Square on both sides 9/4*sin²α=9/4(1-cos²α)=3/4*cos²a+12/5*cosx+48/25 cos²α+4/5*cosα-11/100=0 Alpha is the fourth quadrant cosα>0 cosα=(-4+3√3)/5 Pay attention That's where I don't know where to turn

Merge items of the same kind on both sides and divide by 3

It is known that cos (2 α - β) = negative radical 2 / 2, sin (α - 2 β) = radical 2 / 2, and π / 4

π/4

It is known that sin (3 π - α) = root two cos (3 π / 2 + β), radical three cos (- α) = negative radical two cos (π + β), and 0

Sin (3 π - α) = sin (π - α) = sin α √ 2cos (3 π / 2 + β) = √ 2cos (β - π / 2) = √ 2Sin β. Therefore, the first formula can be transformed into sin α = √ 2Sin β ① √ 3cos (- α) = √ 3cos α - √ 2cos (π + β) = √ 2cos β

It is known that sin (α + Pai / 3) + sin α = - 4 radical 3 / 5, - Pai / 2

Sin (α + Pai / 3) + sin α = - 4 root sign 3 / 5
sinacosπ/3+cosasinπ/3+sina=-4√3/5
3/2sina+√3/2cosa=-4√3/5
√3/2sina+1/2cosa=-4/5
sin(a+π/6)=-4/5