Does $\sin 180+\sin 45=\sin 225$ ?

Hint: Here we have been given an equation and we have to check whether the right hand side value is equal to the left hand side value. Firstly we will take the $LHS$ value and find its value using sine value at the given angles. Then we will take the $RHS$ value and divide the angle in it in two parts and apply the addition formula of trigonometric function sine. Finally we will check whether the $LHS$ is equal to $RHS$ and get our desired answer.

Complete step by step answer:
We have been given the equation as follows:
$\sin 180+\sin 45=\sin 225$
Now we will take $LHS$ function and find its value by using basic values of sine at some angles as follows:
$\Rightarrow LHS=\sin 180+\sin 45$
We know $\sin n\pi =0$ $\forall n\in Z$ and $\sin 45=\dfrac{1}{\sqrt{2}}$ using these values above we get,
$\Rightarrow LHS=0+\dfrac{1}{\sqrt{2}}$
$\Rightarrow LHS=\dfrac{1}{\sqrt{2}}$….$\left( 1 \right)$
Next we will take $RHS$ function and find its value as follows:
$\Rightarrow RHS=\sin 225$
We can write $\sin 225=\sin \left( 180+45 \right)$ so,
$\Rightarrow RHS=\sin \left( 180+45 \right)$
Using the addition formula which is $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ above where $A=180$ and $B=45$ we get,
$\Rightarrow RHS=\sin 180\cos 45+\cos 180\sin 45$
We know that $\sin 180=0,\sin 45=\dfrac{1}{\sqrt{2}},\cos 180=-1,\cos 45=\dfrac{1}{\sqrt{2}}$ substituting the value above we get,
$\Rightarrow RHS=0\times \dfrac{1}{\sqrt{2}}+\left( -1 \right)\times \dfrac{1}{\sqrt{2}}$
$\Rightarrow RHS=-\dfrac{1}{\sqrt{2}}$….$\left( 2 \right)$
On comparing equation (1) and (2) we can see that the two values are not equal.
$\therefore LHS\ne RHS$
Hence the answer is $\sin 180+\sin 45\ne \sin 225$ .

Note:
We should know that angle $225$ lies in the third quadrant and sine in the third quadrant is always negative so from here we can reach the correct answer.