Trigonometric Relation (Tomato Subjective 115)

Problem: If \(\displaystyle { \frac{\sin^4 x }{a} + \frac{\cos^4 x }{b} = \frac{1}{a+b} }\) , then show that \(\displaystyle { \frac{\sin^6 x }{a^2} + \frac{\cos^6 x }{b^2} = \frac{1}{(a+b)^2} }\)

Discussion: 

Put \(\sin^2 x = t \).

The given expression reduces to \(\displaystyle { \frac{t^2 }{a} + \frac{1+t^2 -2t }{b} = \frac{1}{a+b} }\)
\(\displaystyle {\Rightarrow (a+b)t^2 -2at +a – \frac{ab}{a+b} = 0 }\)
\(\displaystyle {\Rightarrow (a+b)^2t^2 -2a(a+b)t +a^2 = 0 }\)
\(\displaystyle {\Rightarrow ((a+b)t-a)^2=0 }\)
\(\displaystyle {\Rightarrow \frac{a}{a+b}=t }\)
\(\displaystyle {\Rightarrow \frac{b}{a+b}=1-t }\)

Hence replacing in the required expression we get

\(\displaystyle { \frac{\sin^6 x }{a^2} + \frac{\cos^6 x }{b^2} }\)
\(\displaystyle {=\frac{t^3 }{a^2} + \frac{(1-t)^3 }{b^2}}\)
\(\displaystyle {=\frac{a^3 }{(a+b)^3a^2} + \frac{b^3 }{(a+b)^3b^2}}\)
\(\displaystyle {=\frac{a}{(a+b)^3} + \frac{b}{(a+b)^3} }\)
\(\displaystyle {=\frac{a+b}{(a+b)^3} }\)
\(\displaystyle {=\frac{1}{(a+b)^2} }\)

(Proved)

Chatuspathi:

  • What is this topic: Trigonometry
  • What are some of the associated concept: Change of Variables
  • Where can learn these topics: Cheenta I.S.I. & C.M.I. course, discusses these topics in the ‘Trigonometry’ module.
  • Book Suggestions: Trigonometry Volume I by S.L. Loney

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