How do you verify the identity #(sin^3beta+cos^3beta)/(sinbeta+cosbeta)=1-sinbetacosbeta#?
1 Answer
see explanation.
Explanation:
There are 3 approaches to verifying the identity.
• Manipulate the left side into the form of the right side
• Manipulate the right side into the form of the left side
• Manipulate both sides until a point is reached where they are
#color(white)(x)"both equal"# Using the first approach.
#"left side "=(sin^3beta+cos^3beta)/(sinbeta+cosbeta)# The numerator is a
#color(blue)"sum of cubes"# and can be factorised in general as follows.
#color(red)(bar(ul(|color(white)(2/2)color(black)(a^3+b^3=(a+b)(a^2+ab+b^2))color(white)(2/2)|)))#
#"here " a=sinbeta" and " b=cosbeta#
#rArrsin^3beta+cos^3beta=#
#=(sinbeta+cosbeta)(sin^2beta-sinbetacosbeta+cos^2beta)# The left side can now be simplified.
#((cancel(sinbeta+cosbeta))(sin^2beta-sinbetacosbeta+cos^2beta))/(cancel(sinbeta+cosbeta))#
#=sin^2beta-sinbetacosbeta+cos^2beta#
#"Using the identity " color(red)(bar(ul(|color(white)(2/2)color(black)(sin^2beta+cos^2beta=1)color(white)(2/2)|)))#
#"Then" sin^2beta-sinbetacosbeta+cos^2beta=1-sinbetacosbeta#
#"Thus left side = right side "rArr"verified"#
