Show that sin 100° − sin 10° is Positive

Solution

We use the identity:

\[ \sin C-\sin D = 2\cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \]

Taking

\[ C=100^\circ, \qquad D=10^\circ \]

Then,

\[ \sin100^\circ-\sin10^\circ = 2\cos\left(\frac{100^\circ+10^\circ}{2}\right) \sin\left(\frac{100^\circ-10^\circ}{2}\right) \]

\[ = 2\cos55^\circ\sin45^\circ \]

Now,

\[ \cos55^\circ>0 \]

and

\[ \sin45^\circ>0 \]

Also,

\[ 2>0 \]

Therefore,

\[ 2\cos55^\circ\sin45^\circ>0 \]

Hence,

\[ \boxed{\sin100^\circ-\sin10^\circ>0} \]

Final Answer

\[ \boxed{\sin100^\circ-\sin10^\circ\text{ is positive}} \]