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

#"see explanation"#

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"#