Known Tan & # 178; a = 2tan & # 178; B + 1 to find cos2a + Sin & # 178; b = how much? Which prawn to solve me, thank you very much

Known Tan & # 178; a = 2tan & # 178; B + 1 to find cos2a + Sin & # 178; b = how much? Which prawn to solve me, thank you very much

We will be able to reach 178; a = 2tan; 178; a = 2tan; 178; a = 2tan; 178; B + 1 so that we can reach 178; a / cos; 178; a = 2Sin; 178; a / cos; 178; a = 2Sin; 178; a; a = 2Sin; 178; a / cos; 178; a = 2Sin; 178; a / cos; 178; a = 2Sin; 178; B + 1sin; 178; B + 1sin; 178; b = 2Sin; 178; bcos; bcos; 178; a + cos; 178; a + cos & \\35;#####; 178; a + cos; 178; ACOS; 178; ACOS; 178; B (1-cos & \\35;35;############a + cos-178; a (1

Tan (a + π / 4) = k, then cos2a=

From the sum angle formula of Tan:
tan(a+π/4)
=(tana+tanπ/4)/(1-tanatanπ/4)
=(1+tana)/(1-tana)
=k
So Tana = (k-1) / (K + 1). Sina / cosa = Tana = (k-1) / (K + 1)
Combining (Sina) ^ 2 + (COSA) ^ 2 = 1, we know (Sina) ^ 2 = (k-1) ^ 2 / (2k ^ 2 + 2)
From the angle doubling formula, we know that cos2a = 1-2 (Sina) ^ 2 = 2K / (k ^ 2 + 1)

If sin2a = a, cos2a = B, then the value of Tan (a + Pai / 4) is

∵ sin2a = a, cos2a = b = = > 2sinacosa = a, 2cos & # 178; A-1 = B (using sine cosine angle doubling formula) = = > 2sinacosa = a, 2cos & # 178; a = B + 1 ∵ Tana = Sina / cosa = (2sinacosa) / (2cos & # 178; a) = A / (B + 1), so tan (a + π / 4) = [...]

If cos2a = m (m does not = 0), then Tan (PAI / 4 + a) =?
Urgent need

∵cos2a=2cosa^2-1=m
｜ cosa = radical [(1 + m) / 2]
Ψ Tana = radical [(1-m) / (1 + m)]
∴tan(pai/4+a)=(tanpai/4+tana)/(1-tanpai/4tana)
=(1 + Tana) / (1-tana) = [1 + radical (1-m ^ 2)] / M

Given Tan (PAI / 4 + a) = 1 / 2, find the value of (sian2a cosa ^ 2) / 1 + cos2a

It is known that: (1 + Tana) / (1-tana) = 1 / 2 = = > Tana = - 1 / 3
∴(sian2a-cosa^2)/1+cos2a=(2sinacosa-cosa^2)/2cosa^2
=(2sina-cosa)/2cosa=tsna-1/2=-5/6

If sin (π / 2-A) = 3 / 5, find the value of COS (π - 2A)?

sin(π/2-a)=cosa=3/5
cos(π-2a)=-cos2a
=-(2cos²a-1)
=-(2×3/5-1)
=-1/5

Given cos (quarter pie + x) = 3 / 5, 17 / 12 π is less than X and 7 / 4 π, find sin 2x of 1-tanx + 2 times sin square X

cos(π/4+x)=3/5,
∴sin2x=-cos[2(π/4+x)]=-[2*(3/5)^2-1]=7/25,
17π/12

Given sin θ = 3 / 5, θ∈ (π / 2, π), then the value of Cos2 θ

Application of double angle formula Cos2 θ = 1-2 (sin θ) ^ 2 = 1-2 * 9 / 25 = 7 / 25

Given that sin α = 5 / 13 and α∈ (Π / 2, Π), the values of Cos2 α, sin α and sin α / 2 are obtained

∵ Sina = 5 / 13, and a ∈ (π / 2, π) ∵ cosa = - √ 1-sina ^ 2 = - √ 1 - (5 / 13) ^ 2 = - 12 / 13 ∵ cos2a = cos ^ 2a-sin ^ a = (12 / 13) ^ 2 - (5 / 13) ^ 2 = 99 / 169sina / 2 = √ (1-cosa) / 2 = √ [1 - (12 / 13)] / 2 = √ 26 / 26cosa / 2 = √ [1 + (12 / 13)] / 2 = 5 √ 26 / 26

Given that sin α = - 1 / 3 and α is the angle of the third quadrant, find cos (α + π / 4), Cos2 α

Sin α = - 1 / 3, and α is the angle of the third quadrant
So cosa is less than 0,
So cosa = - √ (1-sin & # 178; a)
= -√(1-1/9)
= -2√2 /3
From the formula cos (a + b) = cosa * CoSb - Sina * SINB,
cos(α+π/4)=cosa*cos(π/4) -sina*cos(π/4)
= (-2√2 /3) * (√2/2) - (-1/3)*(√2/2)
= -2/3 + √2 /6
And cos2a = 1-2sin & # a
=1- 2*(-1/3)²
=7/9