Question

Find the value of $1+\cos \theta$

Answer 1

Solve the given trigonometric expression

Given, $1+\cos \theta$

We know that, $\mathit{c}\mathit{o}\mathit{s}\left(2\theta \right)\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{-}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }$.

On substituting $\theta$ by $\frac{\theta }{2}$ in the above equation, we get,

$$\begin{gathered} \cos \left(2\times \frac{\theta }{2}\right)={\cos }^2\left(\frac{\theta }{2}\right)-{\sin }^2\left(\frac{\theta }{2}\right) \\ \Rightarrow \cos \left(\theta \right)={\cos }^2\left(\frac{\theta }{2}\right)-{\sin }^2\left(\frac{\theta }{2}\right) \\ \Rightarrow 1+\cos \left(\theta \right)=1+{\cos }^2\left(\frac{\theta }{2}\right)-{\sin }^2\left(\frac{\theta }{2}\right) \\ \Rightarrow 1+\cos \left(\theta \right)=2{\cos }^2\left(\frac{\theta }{2}\right)\mathbf{}\left[\because 1-si{n}^2\theta =co{s}^2\theta \right] \end{gathered}$$

Hence, the value of $1+cos\theta$ is $2co{s}^2\left(\frac{\mathrm{\theta }}{2}\right)$