To prove the identity [tex]\((a \cos A + b \sin A )^2 + (a \sin A - b \cos A )^2 = a^2 + b^2\)[/tex], let's follow a step-by-step approach:
1. Expand [tex]\((a \cos A + b \sin A )^2\)[/tex]:
Calculate the square of the first term.
[tex]\[
(a \cos A + b \sin A)^2 = (a \cos A)^2 + 2(a \cos A)(b \sin A) + (b \sin A)^2
\][/tex]
Simplifying each term we get:
[tex]\[
(a \cos A)^2 = a^2 \cos^2 A
\][/tex]
[tex]\[
2(a \cos A)(b \sin A) = 2ab \cos A \sin A
\][/tex]
[tex]\[
(b \sin A)^2 = b^2 \sin^2 A
\][/tex]
So,
[tex]\[
(a \cos A + b \sin A)^2 = a^2 \cos^2 A + 2ab \cos A \sin A + b^2 \sin^2 A
\][/tex]
2. Expand [tex]\((a \sin A - b \cos A )^2\)[/tex]:
Calculate the square of the second term.
[tex]\[
(a \sin A - b \cos A)^2 = (a \sin A)^2 - 2(a \sin A)(b \cos A) + (b \cos A)^2
\][/tex]
Simplifying each term we get:
[tex]\[
(a \sin A)^2 = a^2 \sin^2 A
\][/tex]
[tex]\[
-2(a \sin A)(b \cos A) = -2ab \sin A \cos A
\][/tex]
[tex]\[
(b \cos A)^2 = b^2 \cos^2 A
\][/tex]
So,
[tex]\[
(a \sin A - b \cos A)^2 = a^2 \sin^2 A - 2ab \sin A \cos A + b^2 \cos^2 A
\][/tex]
3. Combine the two expanded expressions:
Add the two results obtained from the expansions.
[tex]\[
(a \cos A + b \sin A)^2 + (a \sin A - b \cos A)^2
\][/tex]
Which equals:
[tex]\[
(a^2 \cos^2 A + 2ab \cos A \sin A + b^2 \sin^2 A) + (a^2 \sin^2 A - 2ab \sin A \cos A + b^2 \cos^2 A)
\][/tex]
4. Combine like terms:
First, combine [tex]\(a^2 \cos^2 A\)[/tex] and [tex]\(a^2 \sin^2 A\)[/tex]:
[tex]\[
a^2 \cos^2 A + a^2 \sin^2 A = a^2 (\cos^2 A + \sin^2 A)
\][/tex]
Since [tex]\(\cos^2 A + \sin^2 A = 1\)[/tex], this simplifies to [tex]\(a^2\)[/tex].
Next, combine [tex]\(b^2 \sin^2 A\)[/tex] and [tex]\(b^2 \cos^2 A\)[/tex]:
[tex]\[
b^2 \sin^2 A + b^2 \cos^2 A = b^2 (\sin^2 A + \cos^2 A)
\][/tex]
Again, since [tex]\(\sin^2 A + \cos^2 A = 1\)[/tex], this simplifies to [tex]\(b^2\)[/tex].
Lastly, combine [tex]\(2ab \cos A \sin A\)[/tex] and [tex]\(-2ab \sin A \cos A\)[/tex]:
[tex]\[
2ab \cos A \sin A - 2ab \sin A \cos A = 0
\][/tex]
5. Combine the final simplified terms:
[tex]\[
a^2 + b^2 + 0
\][/tex]
Therefore,
[tex]\[
(a \cos A + b \sin A)^2 + (a \sin A - b \cos A)^2 = a^2 + b^2
\][/tex]
We have successfully proven the identity.
