Find the value of $\cos105^\circ+\sin75^\circ$ Find the value of
$$\cos105^\circ+\sin75^\circ.$$
We can write the given trig expression as $$\cos(180^\circ-75^\circ)+\sin75^\circ=-\cos75^\circ+\sin75^\circ\\=\sin75^\circ-\cos75^\circ$$
I don't see what else I can do. Thank you!
 A: $ \sin 75$ can be written as $\sin (30+45)$
$$ \sin (30+45)=\sin 30 \cdot \cos 45 + \cos 30 \cdot \cos 45 $$
Similarly $\cos 105=\cos(60+45) => \cos 60 \cdot \cos 45 - \sin 60 \cdot \sin 45  $
Now just add both the above equation and try to use value from the trignometry chart below.

A: Following on from where you left off,
\begin{align}
\DeclareMathOperator{\deg}{^{\large{\circ}}}
\sin75\deg-\cos75\deg &= -\sin(-75\deg)-\cos(-75\deg) \\[4pt]
&= -\left(\sin(-75\deg)+\cos(-75\deg)\right)
\end{align}
Using the formula $\sin\theta+\cos\theta=\sqrt{2}\sin\left(\theta+45\deg\right)$ (which can be proven using the addition formula for $\sin$), this becomes
$$
-\sqrt{2}\sin(-30\deg) = \sqrt{2}\sin30\deg=\boxed{\frac{\sqrt{2}}{2}} \, .
$$
A: There's some transformation formula I'm listing below,
$\sin (a+b) +  \sin (a-b)= 2 \sin a \cos b$
$\sin (a+b)-\sin (a-b) = 2\cos a \sin b$
$\cos (a+b)+ \cos (a-b) = 2\cos a \cos b$
$\cos (a-b) -\cos (a+b) = 2\sin a \sin b$
Now your problem reduces to $\sin 75° - \cos 75°$, to make use the above formula, we convert $\cos 75°= \sin 15°$ and we write 75 = 45+30 and 15 = 45-30.
So we have $\sin (45°+30°) - \sin (45°-30°) = 2\cos 45° \cdot \sin 30° = 2\cdot \dfrac{1}{√2}\cdot \dfrac{1}{2}= \dfrac{1}{\sqrt{2}}$
Alter: You need not break the given form even, as follows
$\cos 105° + \sin 75°= \cos 105° +\cos 15°= \cos(60+45)+\cos (60-45)$
Check from the above given formulas, and you should get the same answer.
