arccos (x) + arccos(y) = arccos(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)

We will learn how to prove the property of the inverse trigonometric function arccos (x) + arccos(y) = arccos(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)

Proof:  

Let, cos$^{-1}$ x = α and cos$^{-1}$ y = β

From cos$^{-1}$ x = α we get,

x = cos α

and from cos$^{-1}$ y = β we get,

y = cos β

Now, cos (α + β) = cos α cos β - sin α sin β

⇒ cos (α + β) = cos α cos β - $\sqrt{1 - cos^{2} α}$ $\sqrt{1 - cos^{2} β}$

⇒ cos (α + β) = (xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)

⇒ α + β = cos$^{-1}$(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)

⇒ or, cos$^{-1}$ x - cos$^{-1}$ y = cos$^{-1}$(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)

Therefore, arccos (x) + arccos(y) = arccos(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)             Proved.

Note: If x > 0, y > 0 and x$^{2}$ + y$^{2}$ > 1, then the cos$^{-1}$ x + sin$^{-1}$ y may be an angle more than π/2 while cos$^{-1}$(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$), is an angle between – π/2 and π/2.

Therefore, cos$^{-1}$ x + cos$^{-1}$ y = π - cos$^{-1}$(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)

Solved examples on property of inverse circular function arccos (x) + arccos(y) = arccos(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)

1. If cos$^{-1}$$\frac{x}{a}$ + cos$^{-1}$$\frac{y}{b}$ = α prove that,

$\frac{x^{2}}{a^{2}}$ - $\frac{2xy}{ab}$ cos α + $\frac{y^{2}}{b^{2}}$ = sin$^{2}$ α.                        

Solution:  

L. H. S. = cos$^{-1}$$\frac{x}{a}$ + cos$^{-1}$$\frac{y}{b}$ = α

We have, cos$^{-1}$ x - cos$^{-1}$ y = cos$^{-1}$(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)

⇒ cos$^{-1}$ [$\frac{x}{a}$ · $\frac{y}{b}$ - $\sqrt{1 - \frac{x^{2}}{a^{2}}}$ $\sqrt{1 - \frac{y^{2}}{b^{2}}}$] = α

⇒ $\frac{xy}{ab}$ - $\sqrt{(1 - \frac{x^{2}}{a^{2}})(1 - \frac{y^{2}}{b^{2}})}$ = cos α

⇒ $\frac{xy}{ab}$ - cos α = $\sqrt{(1 - \frac{x^{2}}{a^{2}})(1 - \frac{y^{2}}{b^{2}})}$

⇒ ($\frac{xy}{ab}$ - cos α)$^{2}$ = $(1 - \frac{x^{2}}{a^{2}})(1 - \frac{y^{2}}{b^{2}})$, (squaring both the sides)

⇒ $\frac{x^{2}y^{2}}{a^{2}b^{2}}$  - 2$\frac{xy}{ab}$cos α + cos$^{2}$ α = 1 - $\frac{x^{2}}{a^{2}}$ - $\frac{y^{2}}{b^{2}}$ +  $\frac{x^{2}y^{2}}{a^{2}b^{2}}$

⇒ $\frac{x^{2}}{a^{2}}$ - - 2$\frac{xy}{ab}$cos α + cos$^{2}$ α + $\frac{y^{2}}{b^{2}}$ = 1 - cos$^{2}$ α

⇒ $\frac{x^{2}}{a^{2}}$ - - 2$\frac{xy}{ab}$cos α + cos$^{2}$ α + $\frac{y^{2}}{b^{2}}$ = sin$^{2}$ α.            Proved.

2. If cos$^{-1}$ x + cos$^{-1}$ y + cos$^{-1}$ z = π, prove that x$^{2}$ + y$^{2}$ + z$^{2}$ + 2xyz = 1.

Solution:

cos$^{-1}$ x + cos$^{-1}$ y + cos$^{-1}$ z = π

⇒ cos$^{-1}$ x + cos$^{-1}$ y = π - cos$^{-1}$ z

⇒ cos$^{-1}$ x + cos$^{-1}$ y = cos$^{-1}$ (-z), [Since, cos$^{-1}$ (-θ) = π - cos$^{-1}$ θ]

⇒ cos$^{-1}$(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$) = cos$^{-1}$ (-z)

⇒ xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$ = -z

⇒ xy + z = $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$

Now squaring both sides

⇒ (xy + z)$^{2}$ = (1 - x$^{2}$)(1 - y$^{2}$)

⇒ x$^{2}$y$^{2}$ + z$^{2}$ + 2xyz = 1 - x$^{2}$ - y$^{2}$ + x$^{2}$y$^{2}$

⇒ x$^{2}$ + y$^{2}$ + z$^{2}$ + 2xyz = 1                Proved.

● 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

From arccos(x) + arccos(y) to HOME PAGE

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