if sin theta = 2/3 in quadrant 1, find sec theta. HELP!!!
In quadrant 1, both sin and cos are positive.
Therefore for sin(θ)=2/3
cos(θ)=√(1-(2/3)²)
=√(5/9)=√5/3
sec(θ)=3/√5
=3√5/5
howdid you figure that out????
Man this stuff is so hard
If you find this difficult for now, back up a chapter and try the exercises to grasp the building blocks first.
With patience and time, you'll master it eventually.
if cot 17 = 3.2709, what is the tan of 73?
To find sec θ, we need to use the relationship between the trigonometric functions.
To begin, let's recall the definition of secant (sec) in terms of sine (sin):
sec θ = 1 / cos θ
Since we are given that sin θ = 2/3 in quadrant 1, we can use the Pythagorean identity to find cos θ:
sin^2 θ + cos^2 θ = 1
Substituting sin θ = 2/3, we have:
(2/3)^2 + cos^2 θ = 1
4/9 + cos^2 θ = 1
cos^2 θ = 1 - 4/9
cos^2 θ = 5/9
Taking the square root of both sides, we get:
cos θ = √(5/9)
Since we are in quadrant 1 where cosine is positive, the positive square root is taken.
Now we can substitute this value of cos θ into the equation for sec θ:
sec θ = 1 / cos θ
sec θ = 1 / √(5/9)
sec θ = √(9/5)
sec θ = 3/√5
To rationalize the denominator, multiply the numerator and denominator by √5:
sec θ = (3/√5) * (√5/√5)
sec θ = 3√5 / 5
Therefore, sec θ is equal to 3√5/5.