# Is transformation (sinx,cosx) linear?

How can I prove whether $$T(x)=(\sin x,\cos x)$$ is a linear transformation or not?

My work:

$$T:\mathbb R\to R$$ where $$T(x)=(\cos x, \sin x)$$.

$$T(a+b) = T(a) +T(b)\\ \big(\cos(a+b), \sin(a+b) \big) \stackrel{?}{=}\big(\cos a, \sin a\big)+\big(\cos b, \sin b\big)\\ \big(\cos(a+b), \sin(a+b) \big)\ne \big(\cos a+\cos b, \sin a+\sin b\big)$$

NOT linear.

• Please don't use picture and typeset the question using MathJax.
– user65203
Jul 25, 2020 at 14:42

Note in your question that $${T : \mathbb{R}\rightarrow \mathbb{R}^2}$$, not $${T : \mathbb{R}\rightarrow \mathbb{R}}$$. A much simpler way is to look at what happens to "$$0$$" under the transformation. Notice that

$${T(0) = \left(0,1\right)\neq \left(0,0\right)}$$

This automatically means $$T$$ is not a Linear Transformation, since if it were a Linear Transformation we would need the $$0$$ vector in $${\mathbb{R}}$$ (which is just the number $$0$$) to get mapped to the $$0$$ vector in $${\mathbb{R}^2}$$ (which is just $${\left(0,0\right)}$$), but this is not the case.

Your way does also work just fine - just find a concrete example with actual numbers where it doesn't work. You should not just state that $${\sin(a+b)\neq \sin(a) + \sin(b)}$$ - they always want to see a concrete example :) @Yves Daoust has shown you the sort of thing I mean - he plugged in $${a=0}$$ and $${b=0}$$

Short answer: $$\cos(0+0)\ne\cos(0)+\cos(0)$$.

Shorter answer: $$\cos(0)\ne0$$.

(Credit to @Riemann'sPointyNose).