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📐 Mathematics  ·  Inverse Trigonometric Functions  ·  JEE

Find the value of cos[2cos-1(3/5) ] using the double angle formula cos(2theta) = 2cos<sup>2</sup>(theta) - 1.

Answer: -7/25.

  • A -7/25
  • B 7/25
  • C 18/25
  • D -18/25

Correct answer: A. -7/25

Explanation: Let θ=cos⁻¹(3/5), so cosθ=3/5 (θ∈[0,π], principal range). Apply cos2θ=2cos²θ−1=2(9/25)−1=18/25−25/25=−7/25. Since 2θ∈[0,2π], the result −7/25 is valid. Answer: −7/25.

y = sin⁻¹x: Principal Value Branchxyx=-1x=1y=-π/2y=π/2Domain restricted to [-1,1]; range restricted to [-π/2,π/2] - this restriction is what makes the inverse exist

The graph of sin⁻¹x is confined to a narrow domain [-1,1] (since sine itself only takes values in that range) and range [-π/2,π/2] (the principal value branch chosen to make sine one-one and invertible there).

Concept context

Restricting trig functions to make them invertible, the principal value branches of sin-inverse, cos-inverse, tan-inverse, and friends, and the key identities relating them.

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