Thank you STEVE!!!!!
Suppose that Cos (theta) = 1/square root 2
if 0<equal to theta , pi/2 then
sin(theta) =
tan (theta) =
If 3pi/2 less than equal to theta , 2pi then
sin(theta) =
tan (theta) =
I know the trig functions but I feel like I am missing something here. cos (theta) = adj/hyp sin(theta) = opp/hyp...im getting confused with the numbers I am getting..can you help me solve these?
always draw diagrams in the correct quadrants for this kind of stuff.
You know that if cosθ = 1/√2, θ = 45°, and the triangle has sides 1,1,√2
So, in QI, x and y are both positive, and
sinθ = y/r = 1/√2
tanθ = y/x = 1/1 = 1
In QIV, x>0 and y<0, so
sin = y/r = -1/√2
tan = y/x = -1
Of course, I'd be happy to help you understand and solve these trigonometric problems!
Let's start by finding the values of sin(theta) and tan(theta) when 0 ≤ theta ≤ π/2.
Given that cos(theta) = 1/√2, we can use the Pythagorean identity to find sin(theta). The Pythagorean identity states that sin^2(theta) + cos^2(theta) = 1. Therefore, we have:
sin^2(theta) + (1/√2)^2 = 1
sin^2(theta) + 1/2 = 1
To solve for sin(theta), we'll subtract 1/2 from both sides:
sin^2(theta) = 1 - 1/2
sin^2(theta) = 1/2
Taking the square root of both sides gives us:
sin(theta) = ±√(1/2)
Since 0 ≤ theta ≤ π/2, we can conclude that sin(theta) = √(1/2) (positive value) because sin(theta) represents the ratio of the length of the opposite side to the length of the hypotenuse, which is always a positive quantity. Therefore:
sin(theta) = √(1/2) = √2/2
To find tan(theta), we'll use the definition of the tangent function:
tan(theta) = sin(theta) / cos(theta)
Substituting the given values, we get:
tan(theta) = (√2/2) / (1/√2)
tan(theta) = (√2/2) * (√2/1)
tan(theta) = 2/2
tan(theta) = 1
Thus, for 0 ≤ theta ≤ π/2, sin(theta) = √2/2 and tan(theta) = 1.
Now let's move on to the second part of the problem, where 3π/2 ≤ theta ≤ 2π.
Given that cos(theta) = 1/√2, we can use the same approach as before to find sin(theta). Using the Pythagorean identity:
sin^2(theta) + (1/√2)^2 = 1
sin^2(theta) + 1/2 = 1
Subtracting 1/2 from both sides gives us:
sin^2(theta) = 1 - 1/2
sin^2(theta) = 1/2
Taking the square root of both sides gives us:
sin(theta) = ±√(1/2)
However, for 3π/2 ≤ theta ≤ 2π, sin(theta) represents the ratio of the length of the opposite side to the length of the hypotenuse, which is always a negative quantity. Therefore:
sin(theta) = -√(1/2) = -√2/2
To find tan(theta), we'll again use the definition of the tangent function:
tan(theta) = sin(theta) / cos(theta)
Substituting the given values, we get:
tan(theta) = (-√2/2) / (1/√2)
tan(theta) = (-√2/2) * (√2/1)
tan(theta) = -2/2
tan(theta) = -1
Therefore, for 3π/2 ≤ theta ≤ 2π, sin(theta) = -√2/2 and tan(theta) = -1.
I hope this helps clarify the steps and calculations involved in solving these trigonometry problems!
Of course! I'd be happy to help you solve these trigonometric equations.
Let's start with the first condition, which is 0 ≤ θ ≤ π/2.
Given that cos(θ) = 1/√2, we can use the Pythagorean identity to find the value of sin(θ):
sin(θ) = √(1 - cos²(θ))
= √(1 - (1/√2)²)
= √(1 - 1/2)
= √(1/2)
= 1/√2
Therefore, sin(θ) = 1/√2.
Next, let's find the value of tan(θ):
tan(θ) = sin(θ) / cos(θ)
= (1/√2) / (1/√2)
= 1
Therefore, tan(θ) = 1.
Now, let's move on to the second condition, which is 3π/2 ≤ θ ≤ 2π.
Since cos(θ) = 1/√2, we can apply the cosine function's periodicity to find cos(θ):
cos(θ) = cos(θ mod 2π)
θ is already within the range of 3π/2 to 2π, which corresponds to the fourth quadrant on the unit circle. In this quadrant, cos(θ) is positive. Therefore, we can simply use the positive value of 1/√2.
cos(θ) = 1/√2
Now, let's find the value of sin(θ):
sin(θ) = √(1 - cos²(θ))
= √(1 - (1/√2)²)
= √(1 - 1/2)
= √(1/2)
= 1/√2
Thus, sin(θ) = 1/√2.
Finally, let's calculate the value of tan(θ):
tan(θ) = sin(θ) / cos(θ)
= (1/√2) / (1/√2)
= 1
Therefore, tan(θ) = 1.
To summarize:
For 0 ≤ θ ≤ π/2:
sin(θ) = 1/√2
tan(θ) = 1
For 3π/2 ≤ θ ≤ 2π:
sin(θ) = 1/√2
tan(θ) = 1
I hope this explanation helps clarify the process for solving these trigonometric equations. If you have any further questions, please let me know!