Sure, let's prove the given trigonometric identity step-by-step.
We need to prove:
[tex]$(a \cos A + b \sin A )^2+( a \sin A - b \cos A )^2= a ^2+b^2.$[/tex]
Step 1: Expand the left-hand side (LHS).
First, we begin by expanding both squared terms in the LHS:
[tex]\[
(a \cos A + b \sin A )^2 = (a \cos A)^2 + (b \sin A)^2 + 2(a \cos A)(b \sin A)
\][/tex]
[tex]\[
= a^2 \cos^2 A + b^2 \sin^2 A + 2ab \cos A \sin A
\][/tex]
Next, we expand the second term:
[tex]\[
(a \sin A - b \cos A )^2 = (a \sin A)^2 + (-b \cos A)^2 + 2(a \sin A)(-b \cos A)
\][/tex]
[tex]\[
= a^2 \sin^2 A + b^2 \cos^2 A - 2ab \sin A \cos A
\][/tex]
Step 2: Combine the expanded terms.
Now, we combine the expanded terms:
[tex]\[
(a \cos A + b \sin A)^2 + (a \sin A - b \cos A)^2 = (a^2 \cos^2 A + b^2 \sin^2 A + 2ab \cos A \sin A) + (a^2 \sin^2 A + b^2 \cos^2 A - 2ab \sin A \cos A)
\][/tex]
Step 3: Simplify the expression.
Collecting like terms, we get:
[tex]\[
a^2 \cos^2 A + b^2 \sin^2 A + 2ab \cos A \sin A + a^2 \sin^2 A + b^2 \cos^2 A - 2ab \sin A \cos A
\][/tex]
Notice that the [tex]$+2ab \cos A \sin A$[/tex] and [tex]$-2ab \sin A \cos A$[/tex] terms cancel each other out:
[tex]\[
a^2 \cos^2 A + b^2 \sin^2 A + a^2 \sin^2 A + b^2 \cos^2 A
\][/tex]
Combine the [tex]$a^2$[/tex] terms and the [tex]$b^2$[/tex] terms:
[tex]\[
a^2 (\cos^2 A + \sin^2 A) + b^2 (\sin^2 A + \cos^2 A)
\][/tex]
Step 4: Use trigonometric identities.
Recall the fundamental trigonometric identities:
[tex]\[
\sin^2 A + \cos^2 A = 1
\][/tex]
Substitute these identities in the equation:
[tex]\[
a^2 (1) + b^2 (1) = a^2 + b^2
\][/tex]
Thus, we have proven that the left-hand side (LHS) is equal to the right-hand side (RHS):
[tex]\[
(a \cos A + b \sin A )^2+( a \sin A - b \cos A )^2 = a^2 + b^2
\][/tex]
Hence, the identity is proven:
[tex]\[
(a \cos A + b \sin A )^2+( a \sin A - b \cos A )^2 = a^2 + b^2
\][/tex]
