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📐 Mathematics  ·  Inverse Trigonometric Functions  ·  JEE

Evaluate sin[cos-1(4/5) + tan-1(2/3)] using compound angle expansion.

Answer: 17/(5sqrt13).

  • A (8+3sqrt(13))/(5sqrt(13))
  • B (8-3sqrt(13))/(5sqrt(13))
  • C 17/(5sqrt13)
  • D 6/(5sqrt13)

Correct answer: C. 17/(5sqrt13)

Explanation: For cos-1(4/5): cos=4/5, sin=3/5. For tan-1(2/3): in a right triangle with opposite 2, adjacent 3, hyp sqrt13, so sin=2/sqrt13, cos=3/sqrt13. sin(A+B)=sinAcosB+cosAsinB = (3/5)(3/sqrt13)+(4/5)(2/sqrt13) = (9+8)/(5sqrt13) = 17/(5sqrt13).

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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