tan theta= -4/3 and sec theta<0 find sin (theta = pi/6)
If secØ < 0 then cosØ<o
and the cosine is negative in II and III
given tanØ = -4/3, and the tangent is negative in II and IV
so Ø must be in II
I will assume you want to find sin(Ø + π/6) , (the + sign is on the same key as = , you probably did not hit the shift key)
if tanØ = -4/3 in II, then
sinØ = 4/5, and cosØ = -3/5 , (from the 3-4-5 triangle)
sin(Ø+π/6)
= sinØcosπ/6 + cosØsinπ/6
= (4/5)(√3/2) + (-3/5)(1/2)
= (4√3 - 3)/10
To find sin(theta = pi/6), we need to find the value of sin(theta) when theta is pi/6.
First, we are given that tan(theta) = -4/3 and sec(theta) < 0.
From the given information, we can deduce the following:
1. tan(theta) = -4/3. Since tan(theta) = sin(theta)/cos(theta), we can use this to find sin(theta) and cos(theta).
2. sec(theta) < 0. Since sec(theta) = 1/cos(theta), this means that cos(theta) is negative.
Let's solve step-by-step:
Step 1: Finding cos(theta)
We know that tan(theta) = sin(theta)/cos(theta) = -4/3.
Using the Pythagorean identity, sin^2(theta) + cos^2(theta) = 1, we can substitute sin and solve for cos(theta).
Let's assign a value to sin(theta). Since theta = pi/6, which is in the first quadrant, we know that sin(theta) = sin(pi/6) = 1/2 (from the unit circle).
Using sin(theta) = 1/2, we can substitute it into the Pythagorean identity:
(1/2)^2 + cos^2(theta) = 1
1/4 + cos^2(theta) = 1
cos^2(theta) = 3/4
cos(theta) = ±√(3/4)
cos(theta) = ±√3/2
Since sec(theta) = 1/cos(theta), and sec(theta) < 0, we know that cos(theta) is negative. Hence, cos(theta) = -√3/2.
Step 2: Finding sin(theta)
Using the value of cos(theta) = -√3/2, we can substitute it into the Pythagorean identity:
sin^2(theta) + (-√3/2)^2 = 1
sin^2(theta) + 3/4 = 1
sin^2(theta) = 1 - 3/4
sin^2(theta) = 1/4
sin(theta) = ±√(1/4)
sin(theta) = ±1/2
Since theta is in the first quadrant (pi/6 lies between -π/2 and π/2), sin(theta) is positive. Therefore, sin(theta) = 1/2.
So, to answer the question, sin(theta = pi/6) = 1/2.