Evaluation: (1) sin (4 π / 3) cos (7 π / 6) Tan (21 π / 4) (2) sin (23 π / 6) cos (- 17 π / 4) Tan (π / 8) Tan (3 π / 8) requires detailed process

Evaluation: (1) sin (4 π / 3) cos (7 π / 6) Tan (21 π / 4) (2) sin (23 π / 6) cos (- 17 π / 4) Tan (π / 8) Tan (3 π / 8) requires detailed process

(1)sin(4π/3)cos(7π/6)tan(21π/4)=sin(π+π/3)cos(π+π/6)tan(5π+π/4)=-sinπ/3*(-cosπ/6)tanπ/4=-√3/2*(-√3/2)*1=3/4(2)sin(23π/6)cos(-17π/4)tan(π/8)tan(3π/8)=sin(4π-π/6)cos(-4π-π/4)tan(π...

If 3sin θ Tan θ = 8, (0 ＜ θ ＜ π), then cos (θ - π / 4)=

∵0＜θ＜π∴sinθ>0∵3sinθtanθ=8==>3(sinθ)^2/cosθ=8==>(sinθ)^2=8cosθ/3==>(cosθ)^2+8cosθ/3=1 (∵(sinθ)^2+(cosθ)^2=1)==>3(cosθ)^2+8cosθ-3=0==>(3cosθ-1)(cosθ+3)=0==>3cosθ-1=0 (∵cosθ+3>0...

If Tan α = 2, evaluate (sin α - cos α) / (sin α + cos α), sin ^ 2 α + sin α cos α
If Tan α = 2, evaluate (sin α - cos α) / (sin α + cos α), sin ^ 2 α + sin α cos α
Write down the process

tanα=2, sinα/cosα=2, sinα=2cosα,
So (sin α) ^ 2 = 4 (COS α) ^ 2 = 1 - (COS α) ^ 2
We get (COS α) ^ 2 = 1 / 5
(sinα-cosα)/(sinα+cosα)
=(sinα)^2-(cosα)^2=3(cosα)^2=3/5
sin^2α+sinαcosα
=The denominator of (sin ^ 2 α + sin α cos α) / (COS α ^ 2 + sin α ^ 2) is divided by cos α ^ 2
=(tanα ^2+tanα) /(1+tanα ^2)=(4+2)/(1+4)

It is known that Tan (a - β / 2) = 2, Tan (β - A / 2) = - 3. Evaluation: 1. Tan (a + β) / 2 2. Tan (a + β)

1. According to tan (a + b) = (Tana + tanb) / (1-tanatanb), where a is (a - β / 2) and B is (β - A / 2), we can get Tan (A / 2 + β / 2) = - 1 / 7 by substituting;
2. According to tan2a = (2tana) / (1-tanatana), here a is (A / 2 + β / 2), substituting Tan (a + β) = - 7 / 24

If cos (- 100 °) = k, then Tan 80 ° is equal to ()
A. 1−k2kB. −1−k2kC. 1+k2kD. −1+k2k

∵ cos (- 100 °) = k, ∵ cos80 ° = - K, sin80 ° = 1 − cos280 ° = 1 − K2, then tan80 ° = sin80 ° cos80 ° = 1 − K2 − K, so B is selected

Sin [π / 2 + α] = 3 / 5. Find cos 2 α

sin 〔π /2+α〕=cosα =3/5 cos2α=2(cosα)^2-1=2(3/5)^2-1= - 7/25

Given sin (α - β) = 3 / 5, sin (α + β) = - 3 / 5 and (α - β) ∈ (2 / π, π), (α + β) ∈ (3 / 2 π, 2 π), find cos 2 β

cos2β = cos[(α+β)-(α-β)]
=cos(α+β)cos(α-β) + sin(α+β)sin(α-β)
Because (α - β) ∈ (π / 2, π), (α + β) ∈ (3 π / 2,2 π)
Then cos (α - β) = - 4 / 5, cos (α + β) = 4 / 5
Substituting the above formula
cos2β = cos(α+β)cos(α-β) + sin(α+β)sin(α-β)
=-4/5 * 4/5 + (-3/5) * 3/5
=-1

If sin (PAI / 4-A) = 1 / 3, then cos (5 Pai / 4 + a) value?

The original formula = cos [π + π / 4 + a)
=-cos(π/4+a)
=-sin[π/2-(π/4+a)]
=-sin(π/4-a)
=-1/3

Given sin a + cos a = - (1 / 3), the value of sin (PAI + a) + cos (pai-a) is calculated

Sin (PAI + a) + cos (pai-a)
=-sina-cosa
=1/3

Given sin (PAI / 2-A) = 4 / 5, find the value of COS (pai-a)

sin (π/2 - a) = cos a = 4 / 5
So cos (π - a) = - cos a = - 4 / 5