Prove that: \[ \cos20^\circ\cos100^\circ + \cos100^\circ\cos140^\circ – \cos140^\circ\cos20^\circ = -\frac{3}{4} \]
Solution
Use the identity:
\[
\cos A \cos B
=
\frac{1}{2}
\left[
\cos(A+B)+\cos(A-B)
\right]
\]
First term:
\[
\cos20^\circ\cos100^\circ
=
\frac{1}{2}
\left[
\cos120^\circ+\cos(-80^\circ)
\right]
\]
\[
=
\frac{1}{2}
\left[
-\frac{1}{2}+\cos80^\circ
\right]
\]
Second term:
\[
\cos100^\circ\cos140^\circ
=
\frac{1}{2}
\left[
\cos240^\circ+\cos(-40^\circ)
\right]
\]
\[
=
\frac{1}{2}
\left[
-\frac{1}{2}+\cos40^\circ
\right]
\]
Third term:
\[
\cos140^\circ\cos20^\circ
=
\frac{1}{2}
\left[
\cos160^\circ+\cos120^\circ
\right]
\]
\[
=
\frac{1}{2}
\left[
-\cos20^\circ-\frac{1}{2}
\right]
\]
Therefore,
\[
\cos20^\circ\cos100^\circ
+
\cos100^\circ\cos140^\circ
–
\cos140^\circ\cos20^\circ
\]
\[
=
\frac{1}{2}
\left(
-\frac{1}{2}+\cos80^\circ
\right)
+
\frac{1}{2}
\left(
-\frac{1}{2}+\cos40^\circ
\right)
\]
\[
–
\frac{1}{2}
\left(
-\cos20^\circ-\frac{1}{2}
\right)
\]
\[
=
-\frac{1}{4}
+\frac{\cos80^\circ}{2}
-\frac{1}{4}
+\frac{\cos40^\circ}{2}
+\frac{\cos20^\circ}{2}
+\frac{1}{4}
\]
\[
=
-\frac{1}{4}
+
\frac{1}{2}
(\cos20^\circ+\cos40^\circ+\cos80^\circ)
\]
Using the identity:
\[
\cos20^\circ+\cos40^\circ+\cos80^\circ
=
0
\]
Hence,
\[
=
-\frac{1}{4}
+\frac{1}{2}( -1 )
\]
\[
=
-\frac{1}{4}-\frac{1}{2}
=
-\frac{3}{4}
\]
Therefore,
\[
\boxed{
\cos20^\circ\cos100^\circ
+
\cos100^\circ\cos140^\circ
–
\cos140^\circ\cos20^\circ
=
-\frac{3}{4}
}
\]