To show that [tex]\(\cos \left(x+45^{\circ}\right) \cos \left(x-45^{\circ}\right) = \frac{1}{2} \cos 2x\)[/tex], we can utilize trigonometric identities and properties. Here is a step-by-step solution:
1. Expression Setup: We start with the left hand side (LHS) of the equation:
[tex]\[
\cos \left(x + 45^\circ \right) \cos \left(x - 45^\circ \right)
\][/tex]
2. Apply Trigonometric Identity: There is a trigonometric identity that relates the product of cosines to the sum of cosines:
[tex]\[
\cos(A) \cos(B) = \frac{1}{2} \left[ \cos(A + B) + \cos(A - B) \right]
\][/tex]
For our problem, [tex]\(A = x + 45^\circ\)[/tex] and [tex]\(B = x - 45^\circ\)[/tex].
3. Simplify the Angles: Applying the identity with [tex]\(A = x + 45^\circ\)[/tex] and [tex]\(B = x - 45^\circ\)[/tex] gives us:
[tex]\[
\cos \left( x + 45^\circ \right) \cos \left( x - 45^\circ \right) = \frac{1}{2} \left[ \cos \left( (x + 45^\circ) + (x - 45^\circ) \right) + \cos \left( (x + 45^\circ) - (x - 45^\circ) \right) \right]
\][/tex]
4. Combine the Angles: Simplifying inside the cosines:
[tex]\[
\cos \left( (x + 45^\circ) + (x - 45^\circ) \right) = \cos (2x)
\][/tex]
[tex]\[
\cos \left( (x + 45^\circ) - (x - 45^\circ) \right) = \cos (45^\circ + 45^\circ - x + x) = \cos (90^\circ)
\][/tex]
5. Evaluate [tex]\(\cos(90^\circ)\)[/tex]: We know that:
[tex]\[
\cos(90^\circ) = 0
\][/tex]
6. Substitute Back: Substituting the simplified expressions back into the identity we get:
[tex]\[
\cos ( (x + 45^\circ) + (x - 45^\circ) ) = \cos (2x)
\][/tex]
[tex]\[
\cos ( (x + 45^\circ) - (x - 45^\circ) ) = \cos (90^\circ) = 0
\][/tex]
7. Final Simplification: This simplifies our expression to:
[tex]\[
\cos \left( x + 45^\circ \right) \cos \left( x - 45^\circ \right) = \frac{1}{2} \left[ \cos (2x) + 0 \right] = \frac{1}{2} \cos 2x
\][/tex]
Thus, we have shown that:
[tex]\[
\cos \left(x + 45^\circ\right) \cos \left(x - 45^\circ\right) = \frac{1}{2} \cos 2x
\][/tex]
