Solve Equations & Model with Principal Values

We know a special-angle fact:

sin(π/6) = 1/2.

The principal value for arcsin(x) lies in the interval:

−π/2 ≤ x ≤ π/2.

Since π/6 is in this interval and sin(π/6) = 1/2, the principal value is:

x = π/6.

We know cos(π/6) = √3/2.

Cosine is negative in Quadrant II, so the angle with cosine −√3/2 in [0, π] is:

x = π − π/6 = 5π/6.

Because the interval is [0, π], we choose:

x = 5π/6.

We use the special-angle fact:

tan(π/4) = 1.

The principal range for arctan is:

−π/2 < x < π/2.

Since π/4 is in this interval and tan(π/4) = 1, the principal value is:

x = π/4.

In a right triangle, the tangent of the angle at the base is:

tan(θ) = opposite / adjacent.

Here,

opposite = 12 ft (height on the wall)

adjacent = 5 ft (distance from the wall).

So we write:

tan(θ) = 12 / 5

To solve for θ, we use the inverse tangent:

θ = arctan(12/5).

Use the tangent ratio:

tan(θ) = opposite / adjacent = rise / run.

So we write:

tan(θ) = 2 / 8

tan(θ) = 1/4 = 0.25

Now find the angle whose tangent is 0.25:

θ = arctan(0.25) ≈ 14.0°

So the ramp angle is about 14°.

We want the principal value of x such that:

sin(x) = 0.8.

To isolate x, apply the inverse sine:

x = arcsin(0.8).

The principal range for arcsin is:

−π/2 ≤ x ≤ π/2,

and arcsin(0.8) lies in this interval, so the principal value is:

x = arcsin(0.8).

Use the tangent ratio:

tan(θ) = rise / run.

So we write:

tan(θ) = 4 / 12

tan(θ) = 1/3 ≈ 0.3333

Now find the angle whose tangent is 1/3:

θ = arctan(1/3) ≈ 18.4°

So the roof angle is approximately 18.4°.

Start from the equation:

tan(2x) = √3.

We know the special value:

tan(π/3) = √3.

So we set:

2x = π/3

Now divide both sides by 2:

x = π/6

This is the principal value for x.

Use the tangent ratio:

tan(θ) = rise / run.

So we write:

tan(θ) = 30 / 100

tan(θ) = 0.3

Now find the angle whose tangent is 0.3:

θ = arctan(0.3) ≈ 16.7°

So the hill’s incline angle is about 16.7°.

The formula is:

y = A sin(kx).

We are given A = 2 and y = 1, so first substitute:

1 = 2 sin(kx)

Then divide both sides by 2:

sin(kx) = 1/2

Now find the principal value whose sine is 1/2:

kx = arcsin(1/2)

We know sin(π/6) = 1/2, so:

kx = π/6.

Start with the equation:

3cos(x) = 1

Divide both sides by 3:

cos(x) = 1/3

To solve for x, use the inverse cosine:

x = arccos(1/3)

The range of arccos is [0, π], and arccos(1/3) lies in this interval, so this is the principal value.

Use the tangent ratio:

tan(θ) = rise / run.

So we write:

tan(θ) = 50 / 200

tan(θ) = 1/4

tan(θ) = 0.25

Now find the angle whose tangent is 0.25:

θ = arctan(0.25) ≈ 14°

So the climb angle is approximately 14°.

Here the ramp length (10 m) is the hypotenuse, and the rise (2.5 m) is the opposite side.

Use the sine ratio:

sin(θ) = opposite / hypotenuse

So we write:

sin(θ) = 2.5 / 10

sin(θ) = 0.25

Now find the angle whose sine is 0.25:

θ = arcsin(0.25) ≈ 14.5°

So the ramp angle is about 14.5°.

Use the tangent ratio:

tan(θ) = opposite / adjacent.

So we write:

tan(θ) = 25 / 50

tan(θ) = 0.5

Now find the angle whose tangent is 0.5:

θ = arctan(0.5) ≈ 26.6°

So the sun’s angle of elevation is about 26.6°.

Use the tangent ratio:

tan(θ) = rise / run.

So we write:

tan(θ) = 9 / 40

tan(θ) = 0.225

Now find the angle whose tangent is 0.225:

θ = arctan(0.225) ≈ 12.7°

So the slope angle is about 12.7°.

In this right triangle,

opposite side (height on wall) = 7 m

hypotenuse (ladder length) = 8 m

Use the sine ratio:

sin(θ) = opposite / hypotenuse

So we write:

sin(θ) = 7 / 8

To find θ, take the inverse sine:

θ = arcsin(7/8).

We know:

sin(π/6) = 1/2.

A negative sine means the angle is below the x-axis. So we take:

x = −π/6.

Check that it lies in the interval (−π/2, π/2):

−π/6 is between −π/2 and π/2, and

sin(−π/6) = −1/2.

So the principal value is:

x = −π/6.

Use the tangent ratio:

tan(θ) = rise / run.

So we write:

tan(θ) = 10 / 30

tan(θ) = 1/3 ≈ 0.3333

Now find the angle whose tangent is 1/3:

θ = arctan(1/3) ≈ 18.4°

So the slope angle is about 18.4°.

The ratio rise/run = 1/8 means:

tan(θ) = 1 / 8 = 0.125

Now find the angle whose tangent is 0.125:

θ = arctan(0.125) ≈ 7.1°

So the hill’s angle is about 7.1°.

Use the tangent ratio:

tan(θ) = opposite / adjacent.

So we write:

tan(θ) = 40 / 100

tan(θ) = 0.4

Now find the angle whose tangent is 0.4:

θ = arctan(0.4) ≈ 21.8°

So the angle of elevation is about 21.8°.