Question:

If \(\sin(sin^{-1}\frac{1}{5}+\cos^{-1}x)=1\) , then find the value of x.

Updated On: Jan 18, 2024
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Solution and Explanation

\(\sin(sin^{-1}\frac{1}{5}+\cos^{-1}x)=1\)
\(\Rightarrow\sin(\sin^{-1}\frac{1}{5})\cos(\cos^{-1})+\cos(\sin^{-1}\frac{1}{5})\sin(\cos^{-1}x)=1\)
[sin(A+B)=sinAcosB+cosAsinB]
\(\Rightarrow\frac{1}{5}*x+\cos(\sin^{-1}\frac{1}{5})\sin(\cos^{-1}x)=1\)
\(\Rightarrow\frac{1}{5}+\cos(\sin^{-1}\frac{1}{5})\sin(\cos^{-1}x)=1\) .......(1)
Now, let sin-1\((\frac{1}{5})\)=y.
Then ,sin y=\(\frac{1}{5}\) \(\Rightarrow\) cos y= \(\sqrt{1-(\frac{1}{5}^2})=2\sqrt{\frac{6}{5}}\)
\(\Rightarrow\) \(\cos^{-1}(2\frac{\sqrt6}{5})\).
therefore sin-11/5=cos-1(2√6/5)    .......(2)
let cos-1 x=z
Then cosz=x =>sinz=√1-x2
=>z=sin-1(√1-x2)
therefore cos-1x=sin-1(√1-x2)   ....(3)
from (1),(2) and (3)
x/5+cos(cos-1(2√6/5)).sin(sin-1(√1-x2))=1
x/5+2√6/5.√1-x2=1
\(\Rightarrow\) x+2√6 \(\sqrt{1-x^2}\) =5
\(\Rightarrow\) 2√6 \(\sqrt{1-x^2}\) =5-x
On squaring both sides, we get:(4)(6)(1-x2)=25+x2-10x
\(\Rightarrow\) 24-24x2=25+x2-10x
\(\Rightarrow\) 25x2-10x+1=0
\(\Rightarrow\) (5x-1)2=0
\(\Rightarrow\) x=\(\frac{1}{5}\).

Hence, the value of x is \(\frac{1}{5}\)

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Top Questions on Inverse Trigonometric Functions

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Concepts Used:

Properties of Inverse Trigonometric Functions

The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:

Property Set 1:

  • Sin−1(x) = cosec−1(1/x), x∈ [−1,1]−{0}
  • Cos−1(x) = sec−1(1/x), x ∈ [−1,1]−{0}
  • Tan−1(x) = cot−1(1/x), if x > 0  (or)  cot−1(1/x) −π, if x < 0
  • Cot−1(x) = tan−1(1/x), if x > 0 (or) tan−1(1/x) + π, if x < 0

Property Set 2:

  • Sin−1(−x) = −Sin−1(x)
  • Tan−1(−x) = −Tan−1(x)
  • Cos−1(−x) = π − Cos−1(x)
  • Cosec−1(−x) = − Cosec−1(x)
  • Sec−1(−x) = π − Sec−1(x)
  • Cot−1(−x) = π − Cot−1(x)

Property Set 3:

  • Sin−1(1/x) = cosec−1x, x≥1 or x≤−1
  • Cos−1(1/x) = sec−1x, x≥1 or x≤−1
  • Tan−1(1/x) = −π + cot−1(x)

Property Set 4:

  • Sin−1(cos θ) = π/2 − θ, if θ∈[0,π]
  • Cos−1(sin θ) = π/2 − θ, if θ∈[−π/2, π/2]
  • Tan−1(cot θ) = π/2 − θ, θ∈[0,π]
  • Cot−1(tan θ) = π/2 − θ, θ∈[−π/2, π/2]
  • Sec−1(cosec θ) = π/2 − θ, θ∈[−π/2, 0]∪[0, π/2]
  • Cosec−1(sec θ) = π/2 − θ, θ∈[0,π]−{π/2}
  • Sin−1(x) = cos−1[√(1−x2)], 0≤x≤1 = −cos−1[√(1−x2)], −1≤x<0

Property Set 5:

  • Sin−1x + Cos−1x = π/2
  • Tan−1x + Cot−1(x) = π/2
  • Sec−1x + Cosec−1x = π/2

Property Set 6:

  • If x, y > 0

Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1

Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1

  • If x, y < 0

Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1

Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1

Property Set 7:

  • sin−1(x) + sin−1(y) = sin−1[x√(1−y2)+ y√(1−x2)]
  • cos−1x + cos−1y = cos−1[xy−√(1−x2)√(1−y2)]

Property Set 8:

  • sin−1(sin x) = −π−π, if x∈[−3π/2, −π/2]

= x, if x∈[−π/2, π/2]

= π−x, if x∈[π/2, 3π/2]

=−2π+x, if x∈[3π/2, 5π/2] And so on.

  • cos−1(cos x) = 2π+x, if x∈[−2π,−π]

= −x, ∈[−π,0]

= x, ∈[0,π]

= 2π−x, ∈[π,2π]

=−2π+x, ∈[2π,3π]

  • tan−1(tan x) = π+x, x∈(−3π/2, −π/2)

= x, (−π/2, π/2)

= x−π, (π/2, 3π/2)

= x−2π, (3π/2, 5π/2)

Property Set 9: