Question

What is the actual value of arc sin(12/13) +arc sin(4/5)+arc sin(1). show the process using arc sinx +arc sin y formula?


using formula for the first two it comes pi - arc sin (56/65) then + arc sin 1 which is pi/2 . so the final result is 3pi/2 - arc sin (56/65) . and if i dont write pi/2 instead of arc sin 1 , then the result is pi - arc sin (56/65) +arc sin 1 = pi + arc sin (33/65) using formula for arc sin 1 - arc sin (56/65). i m lil bit confused which one is correct ? is there any use of principal value

the calculator shows its approximately 210 degree , means its a third quadrent angle .

normally we know arc sinx +arc sin y = arc sin [x*sqrt(1-y^2) +y*sqrt(1-x^2)]. but if we use this , the answer comes arc sin (56/65) + pi/2 , which does not match with the calculator result . so is there any regulation for arc sinx +arc sin y formula and is it applicable for (arc sinx +arc sin y) < pi/2 ?????

Answer 1

Answer on Question #52550 – Math – Trigonometry

What is the actual value of arc sin(12/13) + arc sin(4/5) + arc sin(1). Show the process using arc sinx + arc sin y formula?

Using formula for the first two it comes pi - arc sin (56/65) then + arc sin 1 which is pi/2. So the final result is 3pi/2 - arc sin (56/65). And if i don't write pi/2 instead of arc sin 1, then the result is pi - arc sin (56/65) + arc sin 1 = pi + arc sin (33/65) using formula for arc sin 1 - arc sin (56/65). i m lil bit confused which one is correct? is there any use of principal value

The calculator shows its approximately 210 degree, means its a third quadrant angle.

Normally we know arc sinx + arc sin y = arc sin [x*sqrt(1-y^2) + y*sqrt(1-x^2)]. But if we use this, the answer comes arc sin (56/65) + pi/2, which does not match with the calculator result. So is there any regulation for arc sinx + arc sin y formula and is it applicable for (arc sinx + arc sin y) < pi/2 ????

Solution

$$\sin^{-1} x + \sin^{-1} y = \left\{ \begin{array}{c} \sin^{-1} \left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right); \, x^2 + y^2 \leq 1 \, \text{or} \, x^2 + y^2 > 1, \, xy < 0 \\ \pi - \sin^{-1} \left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right); \, x^2 + y^2 > 1, \, 0 < x, \, y \leq 1 \\ -\pi - \sin^{-1} \left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right); \, x^2 + y^2 > 1, \, -1 < x, \, y \leq 0 \end{array} \right.$$

Thus,

$$\begin{array}{l} \sin^{-1} \frac{12}{13} + \sin^{-1} \frac{4}{5} = \left| \left( \frac{12}{13} \right)^2 + \left( \frac{4}{5} \right)^2 > 1, \, 0 < \frac{12}{13}, \, \frac{4}{5} \leq 1 \right| = \pi - \sin^{-1} \left( \frac{4}{5} \sqrt{1 - \left( \frac{12}{13} \right)^2} + \frac{12}{13} \sqrt{1 - \left( \frac{4}{5} \right)^2} \right) \\ = \pi - \sin^{-1} \left( \frac{4}{13} + \frac{36}{65} \right) = \pi - \sin^{-1} \left( \frac{56}{65} \right). \end{array}$$

Then

$$\pi - \sin^{-1} \left( \frac{56}{65} \right) + \sin^{-1} 1 = \frac{3\pi}{2} - \sin^{-1} \left( \frac{56}{65} \right) \approx 210{}^\circ.$$

But it really equals

$$\pi + \sin^{-1} \left( \frac{36}{65} \right) \approx 210{}^\circ.$$

So, you saw just trigonometric identity, not a mistake!

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