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.

  • 2
    $\begingroup$ Please don't use picture and typeset the question using MathJax. $\endgroup$
    – user65203
    Jul 25, 2020 at 14:42

2 Answers 2


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).


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