Given that cos (75 ° + θ) = 1 / 3, and θ is the third quadrant angle, the evaluation: cos (- 255 ° - θ) + sin (435 ° + θ)

Given that cos (75 ° + θ) = 1 / 3, and θ is the third quadrant angle, the evaluation: cos (- 255 ° - θ) + sin (435 ° + θ)

Cos (255 ° - θ) + sin (435 ° + θ) = cos (255 ° + θ) + sin (360 ° + 75 ° + θ) = cos (180 ° + 75 ° + θ) + sin (75 ° + θ) = cos (180 ° + 75 ° + θ) + sin (75 ° + θ) θ is the third quadrant angle, 75 ° + θ is the third or fourth quadrant angle, sin (75 ° + θ) = root number (1-cos ^ 2 [(75 ° + θ)]) = = - (2 root number 2) / 3 cos (- 255 ° - θ (- 255 ° - θ (- 255 ° - θ)) = (2 root number 2) / 3 cos (- 255 ° 2) / 3 cos (- 255 ° - θ (- 255 ° - θ)) = (2 root number 2) / 3 cos (2 root number 2) / 3 cos+ sin (435 ° + θ) = - 1 / 3 - (2 root sign 2) / 3

Given that cos (75 ° + θ) = 1 / 3, and θ is the third quadrant angle, find cos (255 ° + θ) + sin (435 ° + θ)

cos(255°+θ)+sin(435°+θ)
=cos(75°+180°+θ)+sin(75°+360°+θ)
=sin(75°+θ)-cos(75°+θ)
θ is the third quadrant angle
cos(75°+θ)=1/3>0
ν 75 ° + θ is the fourth quadrant angle
∴sin(75°+θ)

If cos (75 ° - α) = 1 / 3 and α is the third quadrant angle, then cos (15 ° - α) + sin (α - 15 °) is the value -2 Radix 2 divided by 3

Cos (75 degrees + a) = 1 / 3
=Sin [90 degrees - (75 degrees + a)] = sin (15 degrees - a)
And a is the third quadrant angle,
180 degrees + 2K * 360 degrees

Given that cos (75 ° + α) = 1 / 3, where α is the third quadrant angle, find the value of COS (15 ° - α) + sin (A-15 °)

α is the third quadrant angle
cos(α+75°)>0
The fourth quadrant is α + 75
∵ cos(α+75°)=1/3
∴ sin(α+75°)=-√[1-cos²(α+75°)]=-√(1-1/9)=-2√2/3
∴ cos(15°-α)+sin(α-15°)
=cos(15°-α)-sin(15°-α)
=cos[90°-(75°+α)]-sin[90°-(75°+α)]
=sin(75°+α)-cos(75°+α)
=-2√2/3-1/3
=-(2√2+1)/3

Given that cos (75 ° + α) = 1 / 3 and α is the third quadrant angle, the value of COS (15 ° - α) + sin (α - 15 °) is calculated What I want to know is how to simplify cos (75 ° + α) =?

cos75°cosα+sin75°sinα=1/3

Senior one mathematics: (2008 Shandong college entrance examination) given cos (α - π / 6) + sin α = 4 / 5 √ 3, what is the value of sin (α + 7 / 6 π)?

According to the formula of two angles and formula of trigonofunction, sin (α + 7 / 6 π) = sin (α + 1 / 6 / 6 π) = sin (α) cos (1 / 6 π) + cos (α) sin (1 / 6 π) = 1 / 2 * cos (α + α) + 3 / 3 / 2 * sin (α)). According to the given conditions, cos (α - π / 6) + sin α = cos (α) cos (1 / 6 π) + sin (α) sin (1 / 6 π) + sin (1 / 6 π) + sin α = √ 3 / 2 * cos (α) + cos (α) + α + sin α = √ 3 / 2 * cos (α) + α + (α) + sinα = √ 3 / 2 * cos (α) + α + (α) + α) + 3 / 2 * sin (α) = 4 / 5 √ 3, There is √ 3 * sin (α + 7 / 6 π) = √ 3 / 2 * cos (α) + 3 / 2 * sin (α) = 4 / 5 √ 3. So sin (α + 7 / 6 π) = 4 / 5

Cos (α + β) Let cos (α - β / 2) = - 1 / 9, sin (α / 2 - β) = 2 / 3, α∈ (π / 2, π), β∈ (0, π / 2) There has to be a process, eh

(α - β / 2) - (α / 2 - β) = α / 2 + β / 2. In this way, you first calculate the sine or cosine that gives the difference between the two angles, and pay attention to the range of the angle when square root, and then the angle in the target is twice the angle calculated, and then the formula of double angle of cosine can be used to solve the problem

Evaluation: cos (3 π / 8 - θ) cos (5 π / 24 - θ) + sin (π / 8 + θ) sin (7 π / 24 + θ) Learn the induction formula of trigonometric identities there Evaluation: cos (3 π / 8 - θ) cos (5 π / 24 - θ) + sin (π / 8 + θ) sin (7 π / 24 + θ) I can't count half of it PI is Pai =2cos(3π/8-θ)cos(5π/24-θ) =cos(3π/8-θ-5π/24+θ)+cos(3π/8-θ+5π/24-θ) I just don't know how the two steps change... I did it myself = 2cos (3 π / 8 - θ) cos (5 π / 24 - θ)

Attention relation 3 π / 8 - θ = π / 2 - (π / 8 + θ)
5π/24-θ=π/2-(7π/24+θ)
So the original formula = cos (3 π / 8 - θ) cos (5 π / 24 - θ) + cos (3 π / 8 - θ) sin (7 π / 24 + θ)
=cos(3π/8-θ)cos(5π/24-θ)+cos(3π/8-θ)cos(5π/24-θ)
=2cos(3π/8-θ)cos(5π/24-θ)
=cos(3π/8-θ-5π/24+θ)+cos(3π/8-θ+5π/24-θ)
=cosπ/6+cos(7/12π-2θ)
This is the sum difference formula of product
cosA*cosB=1/2[cos(A-B)+cos(A+B)]

Cos ^ 2 (π / 5) + sin ^ 2 (π / 10)

Cos 2 (π / 5) + sin 2 (π / 10) is used to calculate cos (π / 5), i.e. cos36 ⁰ = 1-2sin? 18 ⁰, and cos36 ⁰ = sin 54 ⁰ = 3sin18 ⁰ - 4sin? 18 ⁰

Given that 0 ＜ α ＜ π / 4, sin (α + π / 4) = 3 / 5, find cos α

0＜α＜π/4
Then π / 4 ＜ α + π / 4 ＜ π / 2
∴cos(α + π/4)＞0
∴cos(α + π/4)=4/5
cosα
=cos[(α + π/4) - π/4]
=cos(α + π/4)cosπ/4 + sin(α + π/4)sinπ/4
=(4/5)×(√2/2) + (3/5)×(√2/2)
=(7/10)√2