• Algebra
• Trigonometry
• Geometry
• Calculus

Find the value of $\sin^{-1} \Big[\sqrt{2} \sin \theta \Big]$ $\,+\,$ $\sin^{-1} \Big[\sqrt{\cos 2\theta}\Big]$

Theta is an angle of a right angled triangle and the two inverse sine functions $\sin^{-1} \Big[\sqrt{2} \sin \theta \Big]$ and $\sin^{-1} \Big[\sqrt{\cos 2\theta}\Big]$ are formed to represents angles but they are connected in addition form to obtain a required result.

$\sin^{-1} \Big[\sqrt{2} \sin \theta \Big]$ $\,+\,$ $\sin^{-1} \Big[\sqrt{\cos 2\theta}\Big]$

Two inverse sine functions are added and the summation of them can be obtained by using the sum law of two inverse sine functions.

$\sin^{-1} x + \sin^{-1} y$ $\,=\,$ $\sin^{-1} [x\sqrt{1-y^2}$ $\,+\,$ $y\sqrt{1-x^2}]$

Apply this rule and simplify the inverse trigonometric expression.

$=\,\,\,$ $\sin^{-1} \Big[\sqrt{2} \sin \theta \sqrt{1-{\Big(\sqrt{\cos 2\theta}\Big)}^2}$ $\,+\,$ $\sqrt{\cos 2\theta}\sqrt{1-{\Big(\sqrt{2} \sin \theta \Big)}^2}\Big]$

$=\,\,\,$ $\sin^{-1} \Big[\sqrt{2} \sin \theta \sqrt{1-\cos 2\theta}$ $\,+\,$ $\sqrt{\cos 2\theta} \sqrt{1-2 \sin^2 \theta} \Big]$

There are two double angle identities in trigonometry to simplify the expression.

Firstly, apply subtraction rule of cosine of double angle from one to express it in terms of square of sine of angle.

$1-\cos 2\theta \,=\, 2\sin^2 \theta$

According to cosine of double angle identity, $1-2 \sin^2 \theta = \cos 2\theta$.

$=\,\,\,$ $\sin^{-1} \Big[\sqrt{2} \sin \theta \sqrt{2\sin^2 \theta}$ $\,+\,$ $\sqrt{\cos 2\theta} \sqrt{\cos 2\theta}\Big]$

$=\,\,\,$ $\sin^{-1} \Big[\sqrt{2} \sin \theta \times \sqrt{2} \sin \theta $ $\,+\,$ $\sqrt{\cos 2\theta} \sqrt{\cos 2\theta}\Big]$

$=\,\,\,$ $\sin^{-1} \Big[{\Big(\sqrt{2} \sin \theta \Big)}^2$ $\,+\,$ ${\Big(\sqrt{\cos 2\theta}\Big)}^2 \Big]$

$=\,\,\,$ $\sin^{-1} \Big[2 \sin^2 \theta \,+\, \cos 2\theta\Big]$

The first trigonometric term inside the inverse sine function is twice the square of sine of angle theta. The second term is cosine of twice the angle theta. The cosine of double angle can be expanded in terms of square of sine of angle. It is good technique which simplifies the inverse trigonometric expression easily.

$\cos 2\theta \,=\, 1-2\sin^2 \theta$

Now, substitute $\cos 2\theta$ by its expansion.

$=\,\,\,$ $\sin^{-1} \Big[2\sin^2 \theta \,+\, 1-2\sin^2 \theta \Big]$

$=\,\,\,$ $\require{cancel} \sin^{-1} \Big[\cancel{2\sin^2 \theta} \,+\, 1-\cancel{2\sin^2 \theta} \Big]$

$=\,\,\,$ $\sin^{-1} \Big[1\Big]$

$=\,\,\,$ $\dfrac{\pi}{2}$

It is the required solution for this inverse trigonometry problem.