Question

Express $\cos \left(2x\right)$ in terms of $\tan \left(x\right)$.

Answer 1

Step 1 : Apply the identity of $\cos \left(2\theta \right)$

As we know the identity $\cos \left(2\theta \right)={\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right)$

so, substituting $\theta =x$,

$\cos \left(2x\right)={\cos }^2\left(x\right)-{\sin }^2\left(x\right)$

Now,

$$\begin{gathered} \cos \left(2x\right)=\frac{{\cos }^2\left(x\right)-{\sin }^2\left(x\right)}{1} \\ \cos \left(2x\right)=\frac{{\cos }^2\left(x\right)-{\sin }^2\left(x\right)}{{\cos }^2\left(x\right)+{\sin }^2\left(x\right)}\left[{\cos }^2\left(x\right)-{\sin }^2\left(x\right)=1\right] \end{gathered}$$

Step 2: Divide the numerator and denominator by ${\cos }^2\left(x\right)$

dividing the numerator and denominator by ${\cos }^2\left(x\right)$

$$\begin{gathered} \cos \left(2x\right)=\frac{{\displaystyle \frac{{\cos }^2\left(x\right)}{{\cos }^2\left(x\right)}}-{\displaystyle \frac{{\sin }^2\left(x\right)}{{\cos }^2\left(x\right)}}}{{\displaystyle \frac{{\cos }^2\left(x\right)}{{\cos }^2\left(x\right)}}+{\displaystyle \frac{{\sin }^2\left(x\right)}{{\cos }^2\left(x\right)}}} \\ \cos \left(2x\right)=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x} \end{gathered}$$

Hence, $\cos \left(2x\right)$in terms of $\tan \left(x\right)$ is $\cos \left(2x\right)=\frac{1-{\tan }^2\left(x\right)}{1+{\tan }^2\left(x\right)}$.