review the proof of\n\\( \\cos ( a - b ) = \\cos a \\cos b + \\sin a \\sin b \\)\nwhich of the following…

review the proof of\n\\( \\cos ( a - b ) = \\cos a \\cos b + \\sin a \\sin b \\)\nwhich of the following complete step 4 of the proof?\n\\( 1 \\) and 1\n2 and 1\n\\( ( \\cos a \\cos b ) ^ { 2 } ( \\sin a \\sin b ) ^ { 2 } \\) and \\( ( \\cos ^ { 2 } ( a - b ) ) ( ( \\sin ^ { 2 } ( a \\)\n- b ) ) )\n\\( ( \\cos ^ { 2 } a + \\sin ^ { 2 } a ) ( \\cos ^ { 2 } b + \\sin ^ { 2 } b ) \\) and \\( ( \\cos ^ { 2 } ( a - b ) ) \\)\n\\( ( \\sin ^ { 2 } ( a - b ) ) ) \nstep 1\n\\( \\sqrt { ( \\cos a - \\cos b ) ^ { 2 } + ( \\sin 4 - 2 \\sin b ) ^ { 2 } } = \\sqrt { ( 2 \\cos ( 4 - b ) - 1 ) ^ { 2 } + ( \\sin ( a - b ) - 0 ) ^ { 2 } }\nstep 2\n\\( ( \\cos a - \\cos b ) ^ { 2 } + ( \\sin a - \\sin b ) ^ { 2 } = ( \\cos ( a - b ) - 1 ) ^ { 2 } + ( \\sin ( a - b ) - 0 ) ^ { 2 }\nstep 3\n\\( \\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} \nstep 4\n\\( \\ - 2 \\cos 4 \\ \\ \\ 2 \\sin 4 \\ \\ \\ = \\ \\ \\ - 2 \\cos ( a - b ) + 1 \nstep 5\n\\( - 2 ( \\cos a \\cos b + \\sin a \\sin b ) = \\cos ( a - b ) \nstep 6

review the proof of\n\\( \\cos ( a - b ) = \\cos a \\cos b + \\sin a \\sin b \\)\nwhich of the following complete step 4 of the proof?\n\\( 1 \\) and 1\n2 and 1\n\\( ( \\cos a \\cos b ) ^ { 2 } ( \\sin a \\sin b ) ^ { 2 } \\) and \\( ( \\cos ^ { 2 } ( a - b ) ) ( ( \\sin ^ { 2 } ( a \\)\n- b ) ) )\n\\( ( \\cos ^ { 2 } a + \\sin ^ { 2 } a ) ( \\cos ^ { 2 } b + \\sin ^ { 2 } b ) \\) and \\( ( \\cos ^ { 2 } ( a - b ) ) \\)\n\\( ( \\sin ^ { 2 } ( a - b ) ) ) \nstep 1\n\\( \\sqrt { ( \\cos a - \\cos b ) ^ { 2 } + ( \\sin 4 - 2 \\sin b ) ^ { 2 } } = \\sqrt { ( 2 \\cos ( 4 - b ) - 1 ) ^ { 2 } + ( \\sin ( a - b ) - 0 ) ^ { 2 } }\nstep 2\n\\( ( \\cos a - \\cos b ) ^ { 2 } + ( \\sin a - \\sin b ) ^ { 2 } = ( \\cos ( a - b ) - 1 ) ^ { 2 } + ( \\sin ( a - b ) - 0 ) ^ { 2 }\nstep 3\n\\( \\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} \nstep 4\n\\( \\ - 2 \\cos 4 \\ \\ \\ 2 \\sin 4 \\ \\ \\ = \\ \\ \\ - 2 \\cos ( a - b ) + 1 \nstep 5\n\\( - 2 ( \\cos a \\cos b + \\sin a \\sin b ) = \\cos ( a - b ) \nstep 6

Answer

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)))