SME
Grade 12 Math · Precalculus
Inverse trigonometric functions
Use this for “Find sin⁻¹(0.5) in degrees and in radians.”
Chart
Domain (blue bar): [−1, 1]
Range (orange bar): [−90°, 90°]
f(0.5) = 30°
This is an example. Type your own values to start a new problem.
sin⁻¹(
x
)
=
A
°
t
The angle in radians
worked out (not typed)
Formulas
A
=
sin⁻¹(
x
)
30 = sin⁻¹(0.5)
t
=
A
×
π
180
π
6
=
30
×
π
180
Assumptions
•
sin⁻¹(x) is the one angle from −90° to 90° whose sine is x.
•
The sine is one-to-one only on −90° to 90°, so the range is restricted there.
•
x must be from −1 to 1: no angle has a sine of 1.5.
•
sin⁻¹(x) is the inverse function, not 1 ÷ sin(x).
Step-by-step
WE KNOW
x = 0.5
FIND
the angle (A), the angle in radians (t)
Step 1 · Find the angle (A)
Use A = sin⁻¹(x)
The
one
angle
from
−90°
to
90°
whose
sine
is
x
.
A
=
sin⁻¹(0.5)
A
=
30°
Step 2 · Find the angle in radians (t)
Use t = A × π/180
A
half
turn,
180°,
is
π
radians:
multiply
by
π
180
.
t
=
30
×
π
180
t
=
π
6
(≈
0.52)
Check
Put the numbers back in. Both sides should match.
sin(30°) = 0.5 ✓
π
6
=
30
×
π
180
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.