Evaluate $\lim_{t\to0}\frac{-(t-2)(\sin t +1)-(t+2)\cos t}{(\sin t + \cos t - 1)t^2}$ without L'hopital 
Question:
Let
  \begin{align}
&S(t):=\int_{\pi/4}^t (\sin t-\sin\left(\frac\pi4\right))dt\\
&T(t):=\frac{\left(\sin t-\sin\left(\frac\pi4\right)\right)\left(t-\frac\pi4\right)}2\\
\end{align}
  Using $$\lim_{t\to0}\frac{\tan t - t}{t^3}=\frac13\tag1$$
  Evaluate the following (without L'hopital)
  \begin{align}
&\quad\lim_{t\to\frac\pi4}\frac{S(t)-T(t)}{T(t)\left(t-\frac\pi4\right)}
\end{align}

What I've done so far is:
$$\lim_{t\to\frac\pi4}\frac{S(t)-T(t)}{T(t)\left(t-\frac\pi4\right)}=\lim_{t\to0}\frac{-\cos\left(t+\frac\pi4\right)+\frac{\sqrt2}2-\left(\sin\left(t+\frac\pi4\right)+\frac{\sqrt2}2\right)\frac t2}{\left(\sin\left(t+\frac\pi4\right)-\frac{\sqrt2}2\right)\frac {t^2}2}$$
$$=\lim_{t\to0}\frac{-(t-2)(\sin t +1)-(t+2)\cos t}{(\sin t + \cos t - 1)t^2}$$
But I don't know how to go from here to the Eq.$(1)$.
Thanks.
 A: This is an extension of my comment. 

Note that if $a=\pi/4$ (this greatly simplifies typing!!!) then $$S(t) =\cos a - \cos t - (t -  a) \sin a, 2T(t)=(t-a)(\sin t - \sin a)$$ therefore $$2S(t)-2T(t)=2(\cos a - \cos t) - (t-a) (\sin a + \sin t) $$ or $$S(t)-T(t)=2\sin\frac{t+a}{2}\sin\frac{t-a}{2}-(t-a)\sin\frac{t+a}{2}\cos\frac{t-a}{2}$$ Let $2u=t+a,2v=t-a$ so that $u\to a, v\to 0$ as $t\to a$. Then we have $$S(t) - T(t) = 2\sin u\cos v(\tan v-v)$$ and hence the desired limit is $$\frac{2}{\cos a} \lim_{t\to a} \frac{S(t) - T(t)} {(t-a) ^{3}}=\frac{1}{2\cos a} \lim_{v\to 0}\sin u\cos v\cdot\frac{\tan v-v}{v^{3}}$$ which is $\dfrac{\tan a} {6}=\dfrac{1}{6}$.
A: Based on your calculations, the given limit and the sum to product formulae,
\begin{align}
  & \lim_{t\to\frac\pi4}\frac{S(t)-T(t)}{T(t)\left(t-\frac\pi4\right)} \\
  =&\lim_{t\to0}\frac{-\cos\left(t+\frac\pi4\right)+\frac{\sqrt2}2-\left(\sin\left(t+\frac\pi4\right)+\frac{\sqrt2}2\right)\frac
  t2}{\left(\sin\left(t+\frac\pi4\right)-\frac{\sqrt2}2\right)\frac
  {t^2}2} \\
  =&
  \lim_{t\to0}\frac{-\cos\left(t+\frac\pi4\right)+\cos(\frac{\pi}{4})-\left(\sin\left(t+\frac\pi4\right)+\sin(\frac{\pi}{4})\right)\frac
  t2}{\left(\sin\left(t+\frac\pi4\right)-\sin(\frac{\pi}{4})\right)\frac
  {t^2}2} \\
  =&
  \lim_{t\to0}\frac{-2\sin\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{-t}{2})-\left(2\sin\left(\frac{t}{2}+\frac\pi4\right)\cos(\frac{t}{2})\right)\frac
  t2}{\left(2\cos\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{t}{2})\right)\frac
  {t^2}2} \\
  =&
  \lim_{t\to0}\frac{2\sin\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{t}{2})-t\sin\left(\frac{t}{2}+\frac\pi4\right)\cos(\frac{t}{2})}
  {t^2\cos\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{t}{2})} \\
  =& \lim_{t\to0} \frac{\sin(\frac{t}{2} +
  \frac{\pi}{4})}{\cos(\frac{t}{2} + \frac{\pi}{4})} \cdot
  \frac{2\sin\left( \frac{t}{2} \right) - t \cos\left(
      \frac{t}{2}
  \right)}{t^2 \sin\left( \frac{t}{2} \right)} \\
  =& \lim_{t\to0} \tan\left(\frac{t}{2} +
  \frac{\pi}{4}\right) \cdot
  \frac{2\sin\left( \frac{t}{2} \right) - t \cos\left(
      \frac{t}{2}
  \right)}{t^2 \sin\left( \frac{t}{2} \right)} \\
  =& \lim_{t\to0} \tan\left(t + \frac{\pi}{4}\right) \cdot
  \frac{2\sin\left( t \right) - 2t \cos\left( t \right)}{(2t)^2 \sin\left( t \right)} \\
  =& \lim_{t\to0} \tan\left(t + \frac{\pi}{4}\right) \cdot
  \frac{\tan\left( t \right) - t}{2t^2 \tan\left( t \right)} \\
  =& \lim_{t\to0} \frac{1}{2} \tan\left(t + \frac{\pi}{4}\right) \cdot
  \frac{\tan\left( t \right) - t}{t^3} \frac{t}{\tan\left( t \right)} \\
  =& \lim_{t\to0} \frac{1}{2} \tan\left(t + \frac{\pi}{4}\right) \cdot
  \frac{\tan\left( t \right) - t}{t^3} \frac{t}{\sin\left( t \right)}
  \cos\left( t \right) \\
  =& \frac{1}{2} \tan\left( 0 + \frac{\pi}{4} \right) \cdot
  \frac{1}{3} \cdot 1 \cdot \cos(0) \\
  =& \frac{1}{6}
\end{align}
