Answer
a) $\alpha =50{}^\circ \text{ and }\beta =5{}^\circ $ in the expansion $\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ $.
b)
The expression $\cos 50{}^\circ \cos 20{}^\circ +\sin 50{}^\circ \sin 20{}^\circ $ is equivalent to $\cos 45{}^\circ $.
c) The exact value of $\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ $ is $\frac{1}{\sqrt{2}}$.
Work Step by Step
(a)
From the difference formula of cosines,
$\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta $
The expansion using the above identity can be written as,
$\cos \left( 50{}^\circ -5{}^\circ \right)=\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ $
Compare the identity with the above expansion to determine the value of $\alpha \text{ and }\beta $.
Hence, $\alpha =50{}^\circ \text{ and }\beta =5{}^\circ $.
(b)
The expansion using the cosine difference formula can be solved as,
$\begin{align}
& \cos \left( 50{}^\circ -5{}^\circ \right)=\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ \\
& \cos 45{}^\circ =\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ
\end{align}$
Hence, the cosine of an angle $45{}^\circ $ is equivalent to the expression $\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ $.
(c)
The expansion using the cosine difference formula can be solved as:
$\begin{align}
& \cos \left( 50{}^\circ -5{}^\circ \right)=\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ \\
& \cos 45{}^\circ =\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ \\
& \frac{1}{\sqrt{2}}=\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ
\end{align}$
Hence, the exact value of $\cos 50{}^\circ \cos 5{}^\circ +\sin 50{}^\circ \sin 5{}^\circ $ is $\frac{1}{\sqrt{2}}$.
