In this revision note, you will learn how to find Principal Value Trigo (Trigonometry) and solve common O-levels A-Math (Additional Math) exam questions.
You will learn:
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In trigonometry, the concept of principal value is essential for solving equations involving trigonometric functions. The principal value refers to the primary value of an inverse trigonometric function within its restricted range. This helps in identifying a specific angle associated with a given trigonometric value, ensuring consistent and accurate solutions. For a more detailed explanation and examples, please refer to the information provided below.
Trigonometric (Trigo) equations can have infinitely many solutions when the interval for the angle x is not specified. For example, consider the graphs of y = cos x and = 1/2 in the figure below, we see there are infinitely many intersections between = cos x and = 1/2. Therefore there would be infinitely many solutions to the equation cos x = 1/2.
If you need to solve for x in the Trigo equation, cos x = 1/2 , you need to use the inverse trigonometric function, y = cos−1(x), to solve for x.
From the graph above, there are infinitely many solutions to the equation cos x = 1/2 .
However, we would obtain only 1 unique solution (which is 60°) from a calculator when we key in cos−1(1/2).
This is also true for sin−1 x and tan−1 x functions and even when negative angles are involved.
For all the inverse trigonometric functions, i.e. sin−1 x, cos−1 x, and tan−1 x function, the calculator will only give a unique value in a specified range.
This value is called the principal value.
The principal value of sin−1 x, where −1 ≤ x ≤ 1, is defined to be the angle in the principal value interval −90° ≤ sin−1 x ≤ 90° (or −π/2 ≤ sin−1 x ≤ π/2). In other words, the principal value of sin−1 x must lie in the 1st or 4th quadrant.
The principal value of cos−1 x, where −1 ≤ x ≤ 1, is defined to be the angle in the principal value interval 0° ≤ cos−1 x ≤ 180° (or 0 ≤ cos−1 x ≤ π). In other words, the principal value of cos−1 x must lie in the 1st or 2nd quadrant.
The principal value of tan−1 x, where x is any real number, is defined to be the angle in the principal value interval –90° ≤ tan−1 x ≤ 90° (or –π/2 ≤ tan−1 x ≤ π/2 ). In other words, the principal value of tan−1 x must lie in the 1st or 4th quadrant.
Step 1: Determine the quadrant of the principal value
The quadrant of the principal value is the common quadrant between the ASTC Trigo rule and the quadrant which the principal value lies in.
Step 2: Use the Special Angle Table
With the trigo ratio and the quadrant, you can refer to the trigo special angle table above to find the principal value.
Without using a calculator, find the principal value of
(i) sin−1(1/2), in degrees,
(ii) cos−1(−1/2) in degrees,
(iii) tan−1(−√3) in radians,
(iv) cos−1(− cos (7π/4) ) in radians
(i) Find the principal value of sin−1(1/2), in degrees,
Let x be the principal value.
Interpretation for sin x =1/2 :
Since sin x is positive, x must lie in the 1st or 2nd quadrant using the ASTC rule.
The principal value interval for sin−1 x is the 1st or 4th quadrant, and −90° ≤sin−1 x ≤ 90°.
So the principal value for sin−1(1/2) must be in the common 1st quadrant.
Referring to the special angles, the acute angle in the 1st quadrant must be 30°.
sin−1(1/2) = 30° (ans)
(ii) Find the principal value of cos−1(−1/2) in degrees,
Let x be the principal value.
Interpretation for cos x = −1/2:
Since cos x is negative, x must lie in the 2nd or 3rd quadrant using the ASTC rule.
The principal value interval for cos−1 x is the 1st or 2nd quadrant, and 0° ≤ cos−1 x ≤ 180°.
So the principal value for cos−1(−1/2) must be in the common 2nd quadrant.
Referring to the table of special angles, the acute angle must be 60°. In the 2nd quadrant, x = 120°
cos−1(−1/2) = 120° (ans)
(iii) Find the principal value of tan −1(−√3) in radians
Let x be the principal value.
Interpretation for tan x = −√3:
Since tan x is negative, x must lie in the 2nd or 4th quadrant using the ASTC Trigo rule.
The principal value interval for tan−1 x is the 1st or 4th quadrant, and −π/2 ≤ tan−1 x ≤ π/2.
So the principal value for tan−1(−√3) must be in the common 4th quadrant.
Referring to the table of special angles, the acute angle must be 60°. In the 4th quadrant, x = −π/3 radians.
tan−1(−√3) = −π/3 (ans)
(iv) Find the principal value of cos−1(−cos (7π/4) ) in radians
Let x be the principal value.
Interpretation:
cos (7π/4) = cos (π/4) (the cosine ratio for angles in the 1st and 4th quadrants are equal).
And cos(π/4) = √2/2 (see table of special angles)
Now cos x = − √2/2.
Since cos x is negative, x must lie in the 2nd or 3rd quadrant using the ASTC rule.
The principal value interval for cos−1 x is the 1st or 2nd quadrant, and 0 ≤ cos−1 x ≤ π .
So the principal value for cos−1(−√2/2) must be in the common 2nd quadrant. Referring to the table of special angles, the acute angle must be π/4. In the 2nd quadrant, x = 3π/4 radians.
cos−1(− cos (7π/4) ) = cos−1(− cos (π/4) ) = cos−1(−√2/2) = 3π/4 (ans)
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