Question

If arc sin x +arc sin y +arc sin z = pi . prove that [x *sqrt(1-x^2)] +[y *sqrt(1-y^2)]+[z *sqrt(1-z^2)] = 2xyz. show the process please

Answer 1

Answer on Question #52565 – Math – Trigonometry

If $arc \sin x + arc \sin y + arc \sin z = \pi$. Prove that $[x\sqrt{1 - x^2}] + [y\sqrt{1 - y^2}] + [z\sqrt{1 - z^2}] = 2xyz$. Show the process please

Solution

Let $arc \sin x = A, arc \sin y = B, arc \sin z = C$. So,

$$A + B + C = \pi.$$

Then

$$\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C.$$

Here is proof of it:

$$\begin{array}{l} \sin 2A + \sin 2B + \sin 2C = 2 \sin (A + B) \cos (A - B) + 2 \sin C \cos C \\ = 2 \sin (\pi - C) \cos (A - B) + 2 \sin C \cos C = 2 \sin (C) \cos (A - B) + 2 \sin C \cos C \\ = 2 \sin (C) (\cos (A - B) + \cos (\pi - (A + B)) = 2 \sin (C) (\cos (A - B) - \cos (A + B)) \\ = 2 \sin (C) (2 \sin B \sin A) = 4 \sin A \sin B \sin C. \end{array}$$ $$2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C = 4 \sin A \sin B \sin C$$

or

$$\sin A \cos A + \sin B \cos B + \sin C \cos C = 2 \sin A \sin B \sin C.$$

We know that

$$\begin{array}{l} \sin A \cos A + \sin B \cos B + \sin C \cos C = \sin A \sqrt{1 - \sin^2 A} + \sin B \sqrt{1 - \sin^2 B} + \sin C \sqrt{1 - \sin^2 C} = \\ = \left[ x \sqrt{1 - x^2} \right] + \left[ y \sqrt{1 - y^2} \right] + \left[ z \sqrt{1 - z^2} \right], \end{array}$$ $$2 \sin A \sin B \sin C = 2xyz.$$

Thus,

$$\left[ x \sqrt{1 - x^2} \right] + \left[ y \sqrt{1 - y^2} \right] + \left[ z \sqrt{1 - z^2} \right] = 2xyz.$$

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