Let $X_{t}=\sum_{i=1}^n(1+s\cdot w)\sin(t_i)$ where $t\in T=[-\pi/2,\pi/2]^n/\{\vec 0\}$, $w\sim\mathbb{N}(0,1)$, $s$ is a scalar denoting the strength of Gaussian noise. How to find the condition on $s$ such that $X_t$ is strictly positive with high probability? i.e. when $n\rightarrow\infty$,
$$P(\inf X_t>0)\rightarrow 1.$$
We have $X_t=Y\sum_{i=1}^n\sin(t_i)$, where $Y:=1+sw$, $t=(t-1,\dots,t_n)\in T^n:=[-\pi/2,\pi/2]^n/\{\vec 0\}$, $w\sim N(0,1)$, and $s$ is a real number.
So, $$\inf_{t\in T^n}X_t=\min_{t\in T^n}X_t=-n|Y|;$$ this minimum is attained at $t=(-\frac\pi2,\dots,-\frac\pi2)$ if $Y\ge0$ and at $t=(\frac\pi2,\dots,\frac\pi2)$ if $Y<0$.
If $s=0$, then $Y=1$. If $s\ne0$, then $Y$ is normal random variable. So, in any case, $P(Y\ne0)=1$.
So, for any real $s$, $$P(\inf_{t\in T^n}X_t>0)=0\not\to1.$$
You can rewrite your equation as
$$\langle \vec 1 + s\cdot \vec g, \vec t\rangle > 0$$
Where $\vec 1$ is the vector of all 1's, $\vec g\sim\mathcal{N}(0, I_n)$ is standard gaussian, and $\vec t := (\sin(t_1),\dots, \sin(t_n))$ is your vector. I will only use that $\vec t$ is in the positive orthant.
We can simplify this equation to
$$\langle \vec 1, \vec t\rangle + s\langle \vec t, \vec g\rangle > 0$$
As $\vec t$ is in the positive orthant, we have that $\langle \vec 1, \vec t\rangle = \lVert \vec t\rVert_1$. It is a standard computation that $s\langle \vec t, \vec g\rangle \sim \mathcal{N}(0,s^2\lVert \vec t\rVert_2^2)$.
Therefore (by symmetry), you reduce to showing the 1D gaussian tail-bound $1 > g'$ for $g'\sim\mathcal{N}(0, s^2\frac{\lVert \vec t\rVert_2^2}{\lVert \vec t\rVert_1^2})$. You should be able to do this with standard bounds, depending on what your standard of "high probability" is.
