Problems on Inverse Trigonometric Function

We will solve different types of problems on inverse trigonometric function.

1. Find the values of sin (cos$^{-1}$ 3/5)

Solution:  

Let, cos$^{-1}$ 3/5 = θ 

Therefore, cos θ = 3/5

Therefore, sin θ = √(1 - cos$^{2}$  θ) = √(1 - 9/25) = √(16/25) = 4/5 .

Therefore, sin (cos$^{-1}$ 3/5) = sin θ = 4/5.

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

Solution:

tan$^{-1}$ sin (- π/2)

= tan$^{-1}$ (- sin π/2)

= tan$^{-1}$ (- 1), [Since - sin π/2 = -1]

= tan$^{-1}$(- tan π/4), [Since tan π/4 = 1]

= tan$^{-1}$ tan (-π/4)

= - π/4. 

Therefore, tan$^{-1}$ sin (- π/2) = - π/4

3. Evaluate: sin$^{-1}$ (sin 10)

Solution:

We know that sin$^{-1}$ (sin θ) = θ, if - $\frac{π}{2}$ ≤ θ ≤ $\frac{π}{2}$.

Here, θ = 10 radians which does not lie between - $\frac{π}{2}$ and $\frac{π}{2}$. But 3π - θ i.e., 3π - 10 lies between - $\frac{π}{2}$ and $\frac{π}{2}$ and sin (3π - 10) = sin 10.

Now, sin$^{-1}$ (sin 10)

= sin^-1 (sin (3π - 10)

= 3π - 10

Therefore, sin$^{-1}$ (sin 10) = 3π - 10.

4. Find the values of cos (tan$^{-1}$ ¾)

Solution:  

Let, tan$^{-1}$ ¾ = θ 

Therefore, tan θ = ¾

We know that sec$^{2}$ θ - tan$^{2}$ θ = 1

⇒ sec θ = √(1 + tan$^{2}$ θ)

⇒ sec θ = √(1 + (3/4)$^{2}$)

⇒ sec θ = √(1 + 9/16)

⇒ sec θ = √(25/16)

⇒ sec θ = 5/4

Therefore, cos θ = 4/5

⇒ θ = cos$^{-1}$ 4/5

Now, cos (tan$^{-1}$ ¾) = cos (cos$^{-1}$ 4/5) = 4/5

Therefore, cos (tan$^{-1}$ ¾) = 4/5

5. Find the values of sec csc$^{-1}$ (2/√3)

Solution:

sec csc$^{-1}$ (2/√3)

= sec csc$^{-1}$ (csc π/3)

= sec (csc$^{-1}$csc π/3)

= sec π/3

= 2

Therefore, sec csc$^{-1}$ (2/√3) = 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

From Problems on Inverse Trigonometric Function to HOME PAGE