sin(θ − ϕ); tan(θ) = 5/12
,
θ in Quadrant III,
sin(ϕ) = − square root 10 over 10
ϕ in Quadrant IV
Please help last one
sin(θ) = -5/13 , cos(θ) = -12/13
cos(ϕ) = .3√10
sin(θ − ϕ) = sin(θ)cos(ϕ) -
... cos(θ)sin(ϕ)
To find the value of sin(θ − ϕ), we can use the trigonometric identity sin(A − B) = sin(A)cos(B) − cos(A)sin(B).
Given that θ is in Quadrant III, we know that the cosine of θ will be negative. However, we need to determine the values of sin(θ) and cos(θ) in order to proceed.
To find sin(θ), we can use the given information that tan(θ) = 5/12. Since tan(θ) = sin(θ)/cos(θ), we can set up the equation:
tan(θ) = sin(θ)/cos(θ)
5/12 = sin(θ)/cos(θ)
Next, we need to use the Pythagorean identity for trigonometric functions in order to find the value of cos(θ). The Pythagorean identity states that sin^2(θ) + cos^2(θ) = 1. Rearranging this equation, we get:
cos^2(θ) = 1 - sin^2(θ)
cos^2(θ) = 1 - (sin(θ))^2
cos(θ) = ± √(1 - (sin(θ))^2)
Since θ is in Quadrant III, the cosine of θ will be negative. So, we can take the negative square root:
cos(θ) = - √(1 - (sin(θ))^2)
Now we can substitute this value of cos(θ) into our equation:
5/12 = sin(θ)/(- √(1 - (sin(θ))^2))
To solve this equation for sin(θ), we need to square both sides:
(5/12)^2 = (sin(θ)/(- √(1 - (sin(θ))^2)))^2
25/144 = sin^2(θ)/(1 - (sin(θ))^2)
Let's call sin(θ) as u for simplicity:
25/144 = u^2/(1 - u^2)
Cross-multiplying, we get:
25(1 - u^2) =144u^2
25 - 25u^2 = 144u^2
25 = 169u^2
u^2 = 25/169
u = ± √(25/169)
Since θ is in Quadrant III, sin(θ) will be negative. So we take the negative square root:
sin(θ) = - √(25/169) = -5/13
Next, we are given that ϕ is in Quadrant IV and sin(ϕ) = -√10/10.
Now we can substitute the values of sin(θ) and sin(ϕ) into the original equation:
sin(θ − ϕ) = sin(θ)cos(ϕ) − cos(θ)sin(ϕ)
sin(θ − ϕ) = (-5/13)(1/√10) − (-√(1 - (sin(θ))^2))( -√10/10)
sin(θ − ϕ) = -5/ (13 * √10) + (-√(1 - (-5/13)^2))( -1/√10)
To simplify the expression further, we need to find the value of cos(θ). We already found cos(θ) = -√(1 - (sin(θ))^2) = -√(1 - (-5/13)^2). We substitute this value into our equation:
sin(θ − ϕ) = -5/ (13 * √10) + (-√(1 - (-5/13)^2))( -1/√10)
sin(θ − ϕ) = -5/ (13 * √10) + (-√(1 - (-5/13)^2))/√10
At this point, we have simplified the expression as much as possible and we have expressed sin(θ − ϕ) in terms of the given information.