Principal Values of Inverse Trigonometric Functions

We will learn how to find the principal values of inverse trigonometric functions in different types of problems.

The principal value of sin$^{-1}$ x for x > 0, is the length of the arc of a unit circle centred at the origin which subtends an angle at the centre whose sine is x. For this reason sin^-1 x is also denoted by arc sin x. Similarly, cos$^{-1}$ x, tan$^{-1}$  x, csc$^{-1}$  x, sec$^{-1}$  x and cot$^{-1}$ x are denoted by arc cos x, arc tan x, arc csc x, arc sec x.

1. Find the principal values of sin$^{-1}$ (- 1/2)      

Solution: 

If θ be the principal value of sin$^{-1}$ x then - $\frac{π}{2}$ ≤ θ ≤ $\frac{π}{2}$.

Therefore, If the principal value of sin$^{-1}$ (- 1/2) be θ then sin$^{-1}$ (- 1/2) = θ

⇒ sin θ = - 1/2 = sin (-$\frac{π}{6}$) [Since, - $\frac{π}{2}$ ≤ θ ≤ $\frac{π}{2}$]

Therefore, the principal value of sin$^{-1}$ (- 1/2) is (-$\frac{π}{6}$).

2. Find the principal values of the inverse circular function cos$^{-1}$ (- √3/2)

Solution:

 If the principal value of cos$^{-1}$ x is θ then we know, 0 ≤ θ ≤ π.

Therefore, If the principal value of cos$^{-1}$  (- √3/2) be θ then cos$^{-1}$  (- √3/2) = θ

⇒ cos θ = (- √3/2) = cos $\frac{π}{6}$ = cos (π - $\frac{π}{6}$) [Since, 0 ≤ θ ≤ π]

Therefore, the principal value of cos$^{-1}$  (- √3/2) is π - $\frac{π}{6}$ = $\frac{5π}{6}$.

3. Find the principal values of the inverse trig function tan$^{-1}$ (1/√3)

Solution:

If the principal value of tan$^{-1}$ x is θ then we know, - $\frac{π}{2}$ < θ < $\frac{π}{2}$.

Therefore, If the principal value of tan$^{-1}$ (1/√3) be θ then tan$^{-1}$ (1/√3) = θ

⇒ tan θ = 1/√3 = tan $\frac{π}{6}$ [Since, - $\frac{π}{2}$ < θ < $\frac{π}{2}$]

Therefore, the principal value of tan$^{-1}$ (1/√3) is $\frac{π}{6}$.

4. Find the principal values of the inverse circular function cot$^{-1}$ (- 1)

Solution:

If the principal value of cot$^{-1}$ x is α then we know, - $\frac{π}{2}$ ≤ θ ≤ $\frac{π}{2}$ and θ ≠ 0.

Therefore, If the principal value of cot$^{-1}$ (- 1) be α then cot$^{-1}$ (- 1) = θ

⇒ cot θ = (- 1) = cot (-$\frac{π}{4}$) [Since, - $\frac{π}{2}$ ≤ θ ≤ $\frac{π}{2}$]  

Therefore, the principal value of cot$^{-1}$ (- 1) is (-$\frac{π}{4}$).    

5. Find the principal values of the inverse trig function sec$^{-1}$ (1)

If the principal value of sec$^{-1}$ x is α then we know, 0 ≤ θ ≤ π and θ ≠ $\frac{π}{2}$.

Therefore, If the principal value of sec$^{-1}$ (1) be α then, sec$^{-1}$ (1) = θ

⇒ sec θ = 1 = sec 0    [Since, 0 ≤ θ ≤ π]

Therefore, the principal value of sec$^{-1}$ (1) is 0.

6. Find the principal values of the inverse trig function csc$^{-1}$ (- 1).

Solution:

If the principal value of csc$^{-1}$ x is α then we know, - $\frac{π}{2}$ ≤ θ ≤ $\frac{π}{2}$ and θ ≠ 0.

Therefore, if the principal value of csc$^{-1}$ (- 1) be θ then csc$^{-1}$ (- 1) = θ

⇒ csc θ = - 1 = csc (-$\frac{π}{2}$) [Since, - $\frac{π}{2}$ ≤ θ ≤ $\frac{π}{2}$]

Therefore, the principal value of csc$^{-1}$ (- 1) is (-$\frac{π}{2}$).

● Inverse Trigonometric Functions

• General and Principal Values of sin$^{-1}$ x
• General and Principal Values of cos$^{-1}$ x
• General and Principal Values of tan$^{-1}$ x
• General and Principal Values of csc$^{-1}$ x
• General and Principal Values of sec$^{-1}$ x
• General and Principal Values of cot$^{-1}$ x
• Principal Values of Inverse Trigonometric Functions
• General Values of Inverse Trigonometric Functions
  
• arcsin(x) + arccos(x) = $\frac{π}{2}$
• arctan(x) + arccot(x) = $\frac{π}{2}$
• arctan(x) + arctan(y) = arctan($\frac{x + y}{1 - xy}$)
• arctan(x) - arctan(y) = arctan($\frac{x - y}{1 + xy}$)
• arctan(x) + arctan(y) + arctan(z)= arctan$\frac{x + y + z – xyz}{1 – xy – yz – zx}$
• arccot(x) + arccot(y) = arccot($\frac{xy - 1}{y + x}$)
• arccot(x) - arccot(y) = arccot($\frac{xy + 1}{y - x}$)
• arcsin(x) + arcsin(y) = arcsin(x $\sqrt{1 - y^{2}}$ + y$\sqrt{1 - x^{2}}$)
• arcsin (x) - arcsin(y) = arcsin (x $\sqrt{1 - y^{2}}$ - y$\sqrt{1 - x^{2}}$)
• arccos (x) + arccos(y) = arccos(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)
• arccos(x) - arccos(y) = arccos(xy + $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)
• 2 arcsin(x) = arcsin(2x$\sqrt{1 - x^{2}}$) 
• 2 arccos(x) = arccos(2x$^{2}$ - 1)
• 2 arctan(x) = arctan($\frac{2x}{1 - x^{2}}$) = arcsin($\frac{2x}{1 + x^{2}}$) = arccos($\frac{1 - x^{2}}{1 + x^{2}}$)
• 3 arcsin(x) = arcsin(3x - 4x$^{3}$)
• 3 arccos(x) = arccos(4x$^{3}$ - 3x)
• 3 arctan(x) = arctan($\frac{3x - x^{3}}{1 - 3 x^{2}}$)
• Inverse Trigonometric Function Formula
• Principal Values of Inverse Trigonometric Functions
• Problems on Inverse Trigonometric Function

11 and 12 Grade Math

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