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Principal values

Prerequisite knowledge

Principal values of $ \sin^{-1} x $:

$ -90^\circ \le \sin^{-1} x \le 90^\circ $

Principal values of $ \cos^{-1} x $:

$ 0^\circ \le \cos^{-1} x \le 180^\circ $

Principal values of $ \tan^{-1} x $:

$ -90^\circ < \tan^{-1} x < 90^\circ $

Special angles:

$30^\circ \text{ or } {\pi \over 6}$ $45^\circ \text{ or } {\pi \over 4}$ $60^\circ \text{ or } {\pi \over 3}$
$\sin \theta$ $ \phantom{.} {1 \over 2} $ $ \phantom{.} {1 \over \sqrt{2}} $ $ \phantom{.} {\sqrt{3} \over 2} $
$\cos \theta$ $ \phantom{.} {\sqrt{3} \over 2} $ $ \phantom{.} {1 \over \sqrt{2}} $ $ \phantom{.} {1 \over 2} $
$\tan \theta$ $ \phantom{.} {1 \over \sqrt{3}} $ $ \phantom{.} 1 $ $ \phantom{.} \sqrt{3} $

Calculating principal values

For $ \sin^{-1} x $:

$$ \boxed{ -90^\circ \le \sin^{-1} x \le 90^\circ } $$


First quadrant

If $x$ is positive, then $ \sin^{-1} x $ lies in the first quadrant such that $ 0^\circ \le \sin^{-1} x \le 90^\circ $.

Fourth quadrant

If $x$ is negative, then $ \sin^{-1} x $ lies in the fourth quadrant such that $ -90^\circ \le \sin^{-1} x \le 0^\circ $.

For $\cos^{-1} x$:

$$ \boxed{ 0^\circ \le \cos^{-1} x \le 180^\circ } $$


First quadrant

If $x$ is positive, then $ \cos^{-1} x $ lies in the first quadrant such that $ 0^\circ \le \cos^{-1} x \le 90^\circ $.

Second quadrant

If $x$ is negative, then $ \cos^{-1} x $ lies in the second quadrant such that $ 90^\circ \le \cos^{-1} x \le 180^\circ $.

For $\tan^{-1} x$:

$$ \boxed{ -90^\circ < \tan^{-1} x < 90^\circ } $$


First quadrant

If $x$ is positive, then $ \tan^{-1} x $ lies in the first quadrant such that $ 0^\circ \le \tan^{-1} x < 90^\circ $.

Fourth quadrant

If $x$ is negative, then $ \tan^{-1} x $ lies in the fourth quadrant such that $ -90^\circ < \tan^{-1} x \le 0^\circ $.

Questions

1. Find, without the use of a calculator, the principal value of sin⁻¹ (-0.5) in radians.

Answer: $ -{\pi \over 6} $

Solutions

2. Without using a calculator, find, in radians, the principal value of $ \sin^{-1} \left( \cos {5 \pi \over 6} \right) $.

Answer: $ -{\pi \over 3} $

Solutions

Past year O level questions

Year & paper Comments
2016 P1 Question 3a State the range of principal values (quite easy)


Trigonometry: R-formulas Differentiation techniques