Question

review the proof of
\\( \cos ( a - b ) = \cos a \cos b + \sin a \sin b \\)
which of the following complete step 4 of the proof?
\\( 1 \\) and 1
2 and 1
\\( ( \cos a \cos b ) ^ { 2 } ( \sin a \sin b ) ^ { 2 } \\) and \\( ( \cos ^ { 2 } ( a - b ) ) ( ( \sin ^ { 2 } ( a \\)

• b ) ) )

\\( ( \cos ^ { 2 } a + \sin ^ { 2 } a ) ( \cos ^ { 2 } b + \sin ^ { 2 } b ) \\) and \\( ( \cos ^ { 2 } ( a - b ) ) \\)
\\( ( \sin ^ { 2 } ( a - b ) ) )
step 1
\\( \sqrt { ( \cos a - \cos b ) ^ { 2 } + ( \sin 4 - 2 \sin b ) ^ { 2 } } = \sqrt { ( 2 \cos ( 4 - b ) - 1 ) ^ { 2 } + ( \sin ( a - b ) - 0 ) ^ { 2 } }
step 2
\\( ( \cos a - \cos b ) ^ { 2 } + ( \sin a - \sin b ) ^ { 2 } = ( \cos ( a - b ) - 1 ) ^ { 2 } + ( \sin ( a - b ) - 0 ) ^ { 2 }
step 3
\\( \begin{array} { r } { \cos ^ { 2 } a - 2 \cos a \cos b + \cos ^ { 2 } b + \sin ^ { 2 } a - 2 \sin a \sin b + \sin ^ { 2 } b } \\ { = \cos ^ { 2 } ( a - b ) - 2 \cos ( a - b ) + 1 + \sin ^ { 2 } ( a - b ) } end{array}
step 4
\\( \\ - 2 \cos 4 \\ \\ \\ 2 \sin 4 \\ \\ \\ = \\ \\ \\ - 2 \cos ( a - b ) + 1
step 5
\\( - 2 ( \cos a \cos b + \sin a \sin b ) = \cos ( a - b )
step 6

Explanation:

Step1: Use the Pythagorean identity

We know that \(\cos^{2}\theta+\sin^{2}\theta = 1\) for any angle \(\theta\).
For the left - hand side of the equation in step 4, \(\cos^{2}A+\sin^{2}A = 1\) and \(\cos^{2}B+\sin^{2}B=1\), so \((\cos^{2}A+\sin^{2}A)(\cos^{2}B+\sin^{2}B)=1\times1 = 1\).

Step2: Simplify the right - hand side of the equation in step 4

We also know that \(\cos^{2}(A - B)+\sin^{2}(A - B)=1\)

Answer:

\((\cos^{2}A+\sin^{2}A)(\cos^{2}B+\sin^{2}B)\) and \((\cos^{2}(A - B)+\sin^{2}(A - B))\)