Given sin + cos divided by sin cos = 2, the value of sin COS is Given sin + cos divided by sin cos = 2, the value of sin times COS is

Given sin + cos divided by sin cos = 2, the value of sin COS is Given sin + cos divided by sin cos = 2, the value of sin times COS is

The solution should be (sin α + cos α) / (sin α - cos α) = 2
The square of both sides is (Sin & # 178; α + cos & # 178; α + 2Sin α cos α) / (Sin & # 178; α + cos & # 178; α - 2Sin α cos α) = 4
That is, (1 + 2Sin α cos α) / (1-2sin α cos α) = 4
That is, 1 + 2Sin α cos α = 4-8sin α cos α
That is 10sin α cos α = 3
That is sin α cos α = 3 / 10

If sin α - sin β = - 1 / 2, cos α + cos β = 1 / 2, then cos (α + β)=______ .

cosα+cosβ=1/2
So (COS α + cos β) ^ 2 = cos ^ α + 2cos α cos β + cos ^ β = 1 / 4
sinα－sinβ =－1/2
(sinα－sinβ )^2=sin^α-2sinαsinβ+sin^β=1/4
Because cos ^ α + sin ^ α = cos ^ β + sin ^ β = 1
So (COS α + cos β) ^ 2 + (sin α - sin β) ^ 2 = 2 + 2cos α cos β - 2Sin α sin β = 1 / 2
2cos(α＋β)=2cosαcosβ-2sinαsinβ=-2+1/2
cos(α＋β)=-3/4

Reduction: Tan (π - a) cos (2 π - a) sin (- A + 3 π / 2) / cos (- A - π) sin (- π - a)

tan(π-a)cos(2π-a)sin(-a+3π/2)/cos(-a-π)sin(-π-a)
=-tanacosa(-cosa)/(-cosasina)
=-1

Reduction of Tan (2 π - α) sin (- 2 π - α) cos (6 π - α) / cos (α - π) sin (5 π - α)

tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-α)
=-tanα(-sinα)cosα/(-cosα)sinα
=-tanα

What is the value of sin6 ^ 2 (Π - α) - cos (α - Π) cos (- α) + 1?

cos(-a)=cosa
cos(a-π)=cos(π-a)=-cosa
sin^2(π-a)=sin^2a
So sin ^ 2 (π - a) - cos (a - π) cos (- a) + 1
=sin^2a+cos^2a+1=2

Evaluation: Tan (2 π - a) cos (3 / 2 π - a) cos (6 π - a) / sin (a + 3 / 2 π) cos (a + 3 / 2 π)

tan（2π-a）cos（3/2π-a）cos（6π-a）/sin（a+3/2π）cos（a+3/2π）=tanacos（-1/2π-a）cos（-a）/sin（a-1/2π）cos（a-1/2π）=-tanacos（1/2π+a）cosa/sin（1/2π-a）cos（1/2π-a）=-tanasinacosa/cosasin...

Given Tan = 2, evaluate
sin²α／1+cos²α

∵tanα=2
∴sin²α／(1+cos²α)
=sin²α／(sin²α+2cos²α)
=(sin²α/cos²α)／[(sin²α/cos²α)+2]
=tan²α/(tan²α+2)
=4/(4+2)
=2/3

Find the value of sin25 π / 6 + cos5 π / 3 + Tan (- 25 π / 4) + sin (- 7 π / 3) × cos (- 13 π / 6) - sin (- 5 π / 6) × cos (- 5 π / 3)

sin25π/6+cos5π/3+tan(-25π/4)+sin(-7π/3)×cos(-13π/6)-sin(-5π/6)×cos(-5π/3)=sin(4π+π/6)+cos(2π-π/3)+tan(-6π-π/4)+sin(-2π-π/3)*cos(-2π-π/6)-sin(-2π+π-π/6)*cos(-2π+π/3)=sin(π/6)+c...

tan(-35π/6)·sin(-46π/3)-cos(37π/6)·tan(55π/6)=?
It's a process

The original formula is Tan (π / 6-6 π) · sin (2 π / 3-16 π) - cos (π / 6 + 6 π) · Tan (π / 6 + 6 π)
=tan(π/6)·sin(2π/3)-cos(π/6)·tan(π/6)
=tan(π/6)·sin(π/3)-cos(π/6)·tan(π/6)
=tan(π/6)·cos(π/6)-cos(π/6)·tan(π/6)
=0

Evaluation: cos (25 π / 4)

=Cos (π / 4) = radical 2 / 2