(a) sin(t)=2/√2 when t=
(b) cos(t)=1/2 when t=
(c) tan(t)=−1 when t=
Steve
answered
45-45-90: sides 1:1:√2
30-60-90: sides 1:√3:2
Each problem here will have two solutions, in the quadrants where the named function is positive or negative.
Explain Bot
answered
(a) sin(t) = 2/√2
To find the exact value of t, we first need to determine the reference angle by taking the inverse sin of 2/√2.
sin^(-1)(2/√2) = π/4
Next, we need to identify the quadrants where sin is positive. Since the given sin(t) = 2/√2 is positive, t must lie in the first and second quadrants.
In the first quadrant (0 ≤ t ≤ π/2), the reference angle is π/4. So, one possible value for t is π/4.
In the second quadrant (π/2 ≤ t ≤ π), the reference angle is also π/4. However, the angle is measured in the clockwise direction. So, the second possible value for t is 3π/4.
Therefore, the solutions for (a) are t = π/4, 3π/4.
(b) cos(t) = 1/2
Similar to part (a), we will start by finding the reference angle by taking the inverse cos of 1/2.
cos^(-1)(1/2) = π/3
Next, we need to identify the quadrants where cos is positive. Since the given cos(t) = 1/2 is positive, t must lie in the first and fourth quadrants.
In the first quadrant (0 ≤ t ≤ π/2), the reference angle is π/3. Therefore, one possible value for t is π/3.
In the fourth quadrant (3π/2 ≤ t ≤ 2π), the reference angle is also π/3. However, the angle is measured in the clockwise direction. So, the second possible value for t is 5π/3.
Therefore, the solutions for (b) are t = π/3, 5π/3.
(c) tan(t) = -1
To find the exact value of t, we will take the inverse tangent of -1.
tan^(-1)(-1) = -π/4
Next, we need to identify the quadrants where tan is negative. Since the given tan(t) = -1 is negative, t must lie in the second and fourth quadrants.
In the second quadrant (π/2 ≤ t ≤ π), the reference angle is π/4. Therefore, one possible value for t is 3π/4.
In the fourth quadrant (3π/2 ≤ t ≤ 2π), the reference angle is also π/4. However, the angle is measured in the clockwise direction. So, the second possible value for t is 7π/4.
Therefore, the solutions for (c) are t = 3π/4, 7π/4.
To summarize:
(a) The solutions are t = π/4, 3π/4.
(b) The solutions are t = π/3, 5π/3.
(c) The solutions are t = 3π/4, 7π/4.
Remember, when dealing with angles, there are infinite solutions due to the periodicity of trigonometric functions. The solutions provided are within the specified interval 0 ≤ t ≤ 2π.
