Question

Show that $\tan \left( \dfrac 12 \sin ^{-1} \dfrac 34 \right) = \dfrac {4-\sqrt{7}}3$ .

Answer 1

Hint: Use the double angle identity of the tangent, $\tan 2 \theta = \dfrac {2 \tan \theta}{1-\tan ^2 \theta}$ to eliminate the troublesome factor of $\dfrac12$ from the argument $\theta = \dfrac 12\sin ^{-1} \dfrac 34$ of the tangent. Then by using the properties of the inverse trigonometric functions, rewrite $\sin^{-1} \dfrac 34$ in terms of other inverse trigonometric functions to absorb the corresponding functions (for instance: $\tan \left( \tan ^{-1} \theta \right) = \theta$ ).

Complete step by step answer:
First of all, we use the well-known double angle identity: $\tan 2 \theta = \dfrac {2a}{1-a^2}$ where $a=\tan \theta \left( \ne \pm 1 \right)$ .
We’ll use this relationship to rewrite this in terms of $a$ . Multiply both sides by $\left(1-a^2\right)$ .
$\left(1-a^2\right)\tan 2 \theta = 2a \implies \tan 2 \theta \cdot a^2 +2 a - \tan 2 \theta =0$
Now, we divide both sides by $\tan 2 \theta$ (if $\neq 0$ ) and use the quadratic formula to solve for $a$ .
${{a}^{2}}+2a\cot 2\theta -1=0\Rightarrow a=\dfrac{-2\cot 2\theta \pm \sqrt{4{{\cot }^{2}}2\theta +4}}{2}$
By using the identity ${{\cot }^{2}}2\theta +1={{\csc }^{2}}2\theta $ , we can simplify this to obtain:
$a=\tan \theta =-\cot 2\theta \pm \csc 2\theta $
For the moment, we consider the equation: $\tan \theta = \csc 2 \theta - \cot 2 \theta$ (both the solutions are valid).
Now, we substitute $\theta = \dfrac 12 \sin ^{-1} \dfrac 34$ in the above equation.
$\tan \left( \dfrac 12 \sin ^{-1} \dfrac 34 \right) = \csc \left( \sin ^{-1} \dfrac 34 \right) - \cot \left( \sin ^{-1} \dfrac 34 \right)$
For simplicity, we assume that
$\tan \left( \dfrac 12 \sin ^{-1} \dfrac 34 \right)=\csc x - \cot y$
We note that if $x={{\sin }^{-1}}\dfrac{3}{4}\Rightarrow \sin x=\dfrac{3}{4}$ then $\csc x=\dfrac{4}{3}\Rightarrow x={{\csc }^{-1}}\dfrac{4}{3}$ .
Moreover, if $y={{\sin }^{-1}}\dfrac{3}{4}\Rightarrow \sin y=\dfrac{3}{4}$ then $\cos y = \sqrt {1-\left(\dfrac34\right)^2}=\dfrac{\sqrt 7}4$ (we may ignore the negative solution if we assume that $y$ lies in the first quadrant).
Hence, $\cot y=\dfrac{\cos y}{\sin y}=\dfrac {\sqrt 7}3$ so that $y={{\cot }^{-1}}\dfrac{\sqrt{7}}{3}$ .
In the end, we substitute the values $x={{\csc }^{-1}}\dfrac{4}{3},\ y={{\cot }^{-1}}\dfrac{\sqrt{7}}{3}$ in the above equation and use the well-known properties of the trigonometric functions to complete the proof.
$\tan ^{-1} \left( \dfrac 12 \sin ^{-1} \dfrac 34 \right) = \csc \left( \csc ^{-1} \dfrac 43 \right) - \cot \left( \cot^{-1} \dfrac {\sqrt 7}3 \right)$
$\tan \left( \dfrac{1}{2}{{\sin }^{-1}}\dfrac{3}{4} \right)=\dfrac{4}{3}-\dfrac{\sqrt{7}}{3}$
Hence, we have shown that $\tan \left( \dfrac{1}{2}{{\sin }^{-1}}\dfrac{3}{4} \right)=\dfrac{4-\sqrt{7}}{3}$
Note: We need to ensure that equations with $\pm$ signs are correctly chosen at each step. For instance, if we choose the equation $\tan \theta=-\csc 2 \theta - \cot 2 \theta$ instead of $\tan \theta=\csc 2 \theta - \cot 2 \theta$ and continue with the solution, we’ll not be able to derive the result quoted in question (though the other result would also be correct because the original equation ${{a}^{2}}+2a\cot 2\theta -1=0$ has two solutions). One way to identify the equation we need to choose is to observe that the result $\dfrac {4-\sqrt{7}}3$ has one positive term and one negative term added together. This is similar to the equation $\tan \theta = \csc 2 \theta - \cot 2 \theta$ where one positive and one negative term are added on the right side (it is always reasonable to assume that an argument i.e. $2\theta $ here lies in the first quadrant as the first choice, however, if we don’t get the correct result then the argument might not lie in the first quadrant and we’ll use the other equation).