How do you prove #(a \sin \theta + b \cos \theta ) ^ { 2} + ( a \cos \theta - b \sin \theta ) ^ { 2} = a ^ { 2} + b ^ { 2}#?
1 Answer
Apr 14, 2018
Explanation:
#"using the "color(blue)"trigonometric identity"#
#•color(white)(x)sin^2theta+cos^2theta=1#
#"consider the left side"#
#"expanding the parenthesis gives"#
#a^2sin^2theta+2ab sinthetacostheta+b^2cos^2thetalarr"left one"#
#a^2cos^2theta-2ab sinthetacostheta+b^2sin^2thetalarr"right one"#
#"summing the 2 expansions gives"#
#a^2sin^2theta+a^2cos^2theta+b^2cos^2theta+b^2sin^2theta#
#=a^2(sin^2theta+cos^2theta)+b^2(cos^2theta+sin^2theta)#
#=a^2+b^2=" right side "rArr" verified"#