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sine of the inverse sine of 1.325

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Marc S. | Polymath tutoring math, science, and writingPolymath tutoring math, science, and wri...
4.9 4.9 (703 lesson ratings) (703)

This is a domain question in disguise. Most high school and college math textbooks introducing inverse trigonometric functions would indicate that the value of sin(sin−1(1.325)) does not exist because sin−1(1.325) doesn't exist.

The sine of any real angle must fall between −1 and 1. This means that the inverse sine function is only permitted to have values between −1 and 1 as its input. Using math vocabulary, the domain of the inverse sine function within the real numbers is [−1,1]. Trying to find sin−1(1.325) within the real numbers is like trying to find √−1 within the real numbers. It doesn't exist because you need a complex number to express the answer.

Of course, if your class is at a more advanced level and is covering the trigonometry of complex numbers then it actually is true that sin(sin−1(1.325)) = 1.325 within complex numbers, but then the question wouldn't be a domain question in disguise.

Carlos D. | Math and Spanish Tutor who caresMath and Spanish Tutor who cares
4.4 4.4 (91 lesson ratings) (91)

Wrong answer.  Sine and inverse sine are  "inverse" functions so they cancel each other...


  The answer (either radians or degrees)   is   1.325

Sorry for the correction,   Carlos


Thanks Carlos!

Bill F. | Experienced Teacher & Tutor in Round Rock, TXExperienced Teacher & Tutor in Round Roc...
5.0 5.0 (1 lesson ratings) (1)

I'm making a few assumptions here:  

1.  Your problem is Inverse Sine of 1.325 radians (not degrees)

2.  By Inverse Sine you also mean Sine-1 or ArcSin

A trig (Trigonometry) table gives ArcSin(1.325 radians) = 14.07°, or x = 0.970