Inverse Trig Functions I
Evaluate arcsin(1)
Answer: π/2
Evaluate sin(2cos(√2/2))
Answer:1
Solve from [0,2π) tan x = -1
Answer: x = 3π/4, 7π/4
What is the midline of the equation f(x)=3 cos(x)-4 ?
Answer: -4
secxcosx-cscxsinx
answer: 0
Evaluate arctan(√3/3)
Answer: π/6
Evaluate cos(arctan 1 - arcsin1)
Answer:√2/2
Solve from (0,2π) sec²x = 4
secx=+/- 2 Answer: (π/3),(2π/3),(4π/3),(5π/3)
What is the period of the function f(t)=sin((π/6)t)?
answer: 12
(cos²x)/(1-sinx)
replace cos²x with (1-sin²x) then factor the difference of squares. Answer: 1+sinx
Evaluate sin(arcsin 1/4)
sin and arcsin cancel each other out therefore... Answer: 1/4
Evaluate tan(arccot x/5)
tan and arccot produce a reciprocal therefore... Answer: 5/x
Solve from (0,2π) sec²(1/2x) = 4
sec(1/2x)=2 or -2... cos(1/2x)=1/2 or-1/2... 1/2x =(π/3),(2π/3),(4π/3),(5π/3)... Answer: (2π/3),(4π/3)
Write an equation for the sinusoidal function that oscillates from a low of -1 to a high of 3, a max at x=-2 and min at x=2.
Answer: -2sin((π/4)t+1
sinx(cotx+tanx)
Write as sin/cos then get a common denominator in the ( ) and distribute sinx. Answer: secx
Evaluate cos(arctan 1)
Answer: √2/2
tan(arcsinx) = ?
Evaluate using a triangle making the side a equal to x and side c equal to 1, therefore making side b equal to √1-x²... Since tan is equal to a/b... Answer: x/√1-x²
Solve from (0,2π) sin²x=sinx
sin²x-sinx=0... sinx(sinx-1)=0... sinx = 0 ||| sinx=1... Answer: x= 0,π ||| x= π/2
A bicycle wheel with radius 14 inches has the bottom-most point on the wheel with a red point. The wheel then begins rolling down the street. Write a formula for the height above ground of the red point after the bicycle has travelled x inches.
The height of the point begins at the lowest value, 0, increases to the highest value of 28 inches, and continues to oscillate above and below a center height of 14 inches. In terms of the angle of rotation, h(θ)= -14cosθ+14
tanx/(1+secx) + (1+secx)/tanx
Find a common denominator. Answer: 2cscx
Evaluate sec(arccos -√3/2)
Answer: (-2√3)/3
tan(arcsec x/2) = ?
Solve using a triangle making side a equal 2 and side c equal x...After using Pythagorean theorem to get side b we know that the solution is the equivalent of b/a therefore... Answer: √x²-4/2
Solve from [0,2π) 4sin²[(π/3)x]=3
sin[u] =+/- √3/2 then u = (π/3 +πn),(2π/3+πn). Now replace u with (π/3)x and solve for x. Answer: x=1,2,4,5
A point completes 1 revolution every 2 minutes around circle of radius 5. Find the x coordinate of the point as a function of time.
x=r cos (theta) so x(t) = 5cos (theta)
sin^4(x)-cos^4(x)
Factor the difference of squares then replace cos²x with 1-sin²x. Answer: 2sin²x-1