Inverse Trigonometric Functions: Diagram-Based Questions (5)

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Tags Inverse Trigonometric Functions

Question

Evaluate sin⁡ ⁣[sin⁡−1 ⁣35+cos⁡−1 ⁣1213]\sin\!\left[\sin^{-1}\!\tfrac{3}{5} + \cos^{-1}\!\tfrac{12}{13}\right].

Solution — Step by Step

Let α=sin⁡−135\alpha = \sin^{-1}\tfrac{3}{5}. Then sin⁡α=3/5\sin\alpha = 3/5, so think of a right triangle with opposite =3= 3, hypotenuse =5= 5, adjacent =4= 4. Hence cos⁡α=4/5\cos\alpha = 4/5.

Let β=cos⁡−11213\beta = \cos^{-1}\tfrac{12}{13}. Then cos⁡β=12/13\cos\beta = 12/13, with right triangle adjacent =12= 12, hypotenuse =13= 13, opposite =5= 5. Hence sin⁡β=5/13\sin\beta = 5/13.

 sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta sin⁡(α+β)=35⋅1213+45⋅513\sin(\alpha + \beta) = \tfrac{3}{5} \cdot \tfrac{12}{13} + \tfrac{4}{5} \cdot \tfrac{5}{13} =3665+2065=5665= \tfrac{36}{65} + \tfrac{20}{65} = \tfrac{56}{65}

Final answer: 5665\mathbf{\tfrac{56}{65}}.

Why This Works

Inverse trig values are angles, and we already know how to manipulate angles using sum and difference formulas. The trick is to convert the inverse trig expression into the sine/cosine of an angle using a right triangle — then everything reduces to ordinary trigonometry.

The right-triangle visualization is fast and avoids errors. It also makes principal-value issues disappear: as long as the value is in the principal range, the triangle approach is valid.

Alternative Method

Use the formula sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x + \cos^{-1}x = \pi/2 to swap functions if needed. Or: use sin⁡−1x+sin⁡−1y=sin⁡−1 ⁣(x1−y2+y1−x2)\sin^{-1}x + \sin^{-1}y = \sin^{-1}\!\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) — but only when the result lies in [−π/2,π/2][-\pi/2, \pi/2] and certain inequalities hold. The triangle approach is more bullet-proof.

Common Mistake

Treating sin⁡−1\sin^{-1} as 1/sin⁡1/\sin — a notational confusion. sin⁡−1(x)\sin^{-1}(x) is the inverse function (angle whose sine is xx), while 1/sin⁡(x)=csc⁡(x)1/\sin(x) = \csc(x). The notation arcsin⁡\arcsin removes the ambiguity but isn’t standard in CBSE.

Memorise the principal-value ranges: sin⁡−1:[−π/2,π/2]\sin^{-1}: [-\pi/2, \pi/2], cos⁡−1:[0,π]\cos^{-1}: [0, \pi], tan⁡−1:(−π/2,π/2)\tan^{-1}: (-\pi/2, \pi/2). JEE Main loves to give a value outside the principal range and ask you to “fix” it — for example, sin⁡−1(sin⁡4)≠4\sin^{-1}(\sin 4) \neq 4 because 4>π/24 > \pi/2.

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