Question

What is $\sin \left(x\right)+\cos \left(x\right)$in terms of $\mathrm{sine}$?

Answer 1

Transformation of cosx into sinx:

By using complementary angles identities,

$\cos \left(x\right)=\sin \left(\frac{\mathrm{\pi }}{2}-x\right)$

$\therefore \sin \left(x\right)+\cos \left(x\right)=\sin \left(x\right)+\sin \left(\frac{\mathrm{\pi }}{2}-x\right)$

By using the trigonometric identity,

$$\begin{gathered} \therefore {\sin }^{2}x+{\cos }^{2}x=1 \\ \Rightarrow \cos x=\sqrt{1-{\sin }^{2}x} \end{gathered}$$

$\therefore \sin \left(x\right)+\cos \left(x\right)=\sin \left(x\right)+\sqrt{1-{\sin }^{2}x}$

Hence, $\mathbf{}\mathbf{sin}\mathbf{}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{sin}\mathbf{}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}\mathbf{sin}\mathbf{}\left(\frac{\pi }{2}-x\right)$

$\mathbf{}\mathbf{sin}\mathbf{}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{sin}\mathbf{}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{sin}}^{\mathbf{2}}\mathbf{}\mathbf{x}}$