Question

How do you write $\tan \left( 2x \right)$ in terms of $\sin x$?

Answer 1

To solve the above question, we need to use the trigonometric identity of the tangent function given by $\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$. Then, since according to the above question we have to write $\tan \left( 2x \right)$ in terms of $\sin x$, we will substitute $\tan x=\dfrac{\sin x}{\cos x}$ into the identity $\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$ so as to get $\tan 2x=\dfrac{2\sin x\cos x}{{{\cos }^{2}}x-{{\sin }^{2}}x}$. Then we have to substitute the trigonometric identity ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ into the numerator and the trigonometric identity $\cos x=\sqrt{1-{{\sin }^{2}}x}$ into the denominator, the expression for $\tan \left( 2x \right)$ will get simplified and we will obtain the value of $\tan \left( 2x \right)$ in terms of $\sin x$.

Complete step by step answer:
We know that the trigonometric identity for $\tan \left( 2x \right)$ is given by
$\Rightarrow \tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$
Now, we know that $\tan x=\dfrac{\sin x}{\cos x}$. On substituting this in the above identity, we get

& \Rightarrow \tan 2x=\dfrac{2\dfrac{\sin x}{\cos x}}{1-{{\left( \dfrac{\sin x}{\cos x} \right)}^{2}}} \\\ & \Rightarrow \tan 2x=\dfrac{2\dfrac{\sin x}{\cos x}}{\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x}} \\\ & \Rightarrow \tan 2x=\dfrac{2\dfrac{\sin x}{\cos x}{{\cos }^{2}}x}{{{\cos }^{2}}x-{{\sin }^{2}}x} \\\ \end{aligned}$$ On simplifying the numerator, we get $$\Rightarrow \tan 2x=\dfrac{2\sin x\cos x}{{{\cos }^{2}}x-{{\sin }^{2}}x}........\left( i \right)$$ Now, we know the trigonometric identity which is given by $\Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x=1$ On subtracting ${{\sin }^{2}}x$ from both the sides of the above equation, we get $\begin{aligned} & \Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x-{{\sin }^{2}}x=1-{{\sin }^{2}}x \\\ & \Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x........\left( ii \right) \\\ \end{aligned}$ Now, on taking the square root on both the sides of the above equation, we get $\Rightarrow \cos x=\sqrt{1-{{\sin }^{2}}x}........\left( iii \right)$ Now, we substitute the above equations (ii) and (iii) into the identity (i) to get $$\begin{aligned} & \Rightarrow \tan 2x=\dfrac{2\sin x\sqrt{1-{{\sin }^{2}}x}}{1-{{\sin }^{2}}x-{{\sin }^{2}}x} \\\ & \Rightarrow \tan 2x=\dfrac{2\sin x\sqrt{1-{{\sin }^{2}}x}}{1-2{{\sin }^{2}}x} \\\ \end{aligned}$$ **Hence, we have finally obtained $\tan \left( 2x \right)$ in terms of $\sin x$ as $$\dfrac{2\sin x\sqrt{1-{{\sin }^{2}}x}}{1-2{{\sin }^{2}}x}$$.** **Note:** For solving these types of questions, we must remember different trigonometric identities, the double angle formulae for the tangent, sine and cosine functions. We can also write $\tan \left( 2x \right)$ as $\dfrac{\sin \left( 2x \right)}{\cos \left( 2x \right)}$ and then substitute the identity $\sin 2x=2\sin x\cos x$ into the numerator and the identity $\cos 2x=1-2{{\sin }^{2}}x$ into the denominator to directly get $\tan \left( 2x \right)=\dfrac{2\sin x\cos x}{1-2{{\sin }^{2}}x}$. Then on substituting $\cos x=\sqrt{1-{{\sin }^{2}}x}$ into the numerator, we will obtain the final expression.