text: # Trig
  
  • May 6th 2012, 05:06 AM
  srirahulan
  Trig
  a=2cosA+3sinA , b=3cosA+2sinA So give a relation between (a) and (b) without A.
  • May 6th 2012, 05:39 AM
  Prove It
  Re: Trig
  \displaystyle \begin{align*} a &= 2\cos{A} + 3\sin{A} \\ b &= 3\cos{A} + 2\sin{A} \\ \\ 3a &= 6\cos{A} + 9\sin{A} \\ 2b &= 6\cos{A} + 4\sin{A} \\ \\ 3a - 2b &= 6\cos{A} + 9\sin{A} - \left(6\cos{A} + 4\sin{A}\right) \\ 3a - 2b &= 5\sin{A} \\ \frac{3}{5}a - \frac{2}{5}b &= \sin{A} \\ \left(\frac{3}{5}a - \frac{2}{5}b\right)^2 &= \sin^2{A} \\ \frac{9}{25}a^2 - \frac{12}{25}ab + \frac{4}{25}b^2 &= \sin^2{A} \\ \\ 2a &= 4\cos{A} + 6\sin{A} \\ 3b &= 9\cos{A} + 6\sin{A} \\ \\ 3b - 2a &= 9\cos{A} + 6\sin{A} - \left(4\cos{A} + 6\sin{A}\right) \\ 3b - 2a &= 5\cos{A} \\ \frac{3}{5}b - \frac{2}{5}a &= \cos{A} \\ \left(\frac{3}{5}b - \frac{2}{5}a\right)^2 &= \cos^2{A} \\ \frac{9}{25}b^2 - \frac{12}{25}ab + \frac{4}{25}a^2 &= \cos^2{A} \end{align*}
  
  \displaystyle \begin{align*} \frac{9}{25}a^2 - \frac{12}{25}ab + \frac{4}{25}b^2 + \frac{9}{25}b^2 - \frac{12}{25}ab + \frac{4}{25}a^2 &= \sin^2{A} + \cos^2{A} \\ \frac{13}{25}a^2 - \frac{24}{25}ab + \frac{13}{25}b^2 &= 1 \\ a^2 - \frac{24}{13}ab + b^2 &= \frac{25}{13} \\ a^2 - \frac{24}{13}ab + \left(-\frac{12}{13}b\right)^2 - \left(-\frac{12}{13}b\right)^2 + b^2 &= \frac{25}{13} \\ \left(a - \frac{12}{13}b\right)^2 - \frac{144}{169}b^2 + \frac{169}{169}b^2 &= \frac{325}{169} \\ \left(a - \frac{12}{13}b\right)^2 + \frac{25}{169}b^2 &= \frac{325}{169} \\ \left(a - \frac{12}{13}b\right)^2 &= \frac{325 - 25b^2}{169} \\ a - \frac{12}{13}b &= \frac{\pm \sqrt{325 - 25b^2}}{13} \\ a &= \frac{12b \pm \sqrt{325 - 25b^2}}{13}\end{align*}
  • May 6th 2012, 05:54 AM
  skeeter
  Re: Trig
  Quote:
  
  Originally Posted by srirahulan
  a=2cosA+3sinA , b=3cosA+2sinA So give a relation between (a) and (b) without A.
  
  $3a = 6\cos{A}+9\sin{A}$
  $-2b = -6\cos{A}-4\sin{A}$
  -----------------------------------------
  $3a-2b = 5\sin{A}$
  
  $\sin{A} = \frac{3a-2b}{5}$
  
  $3b = 9\cos{A}+6\sin{A}$
  $-2a=-4\cos{A}-6\sin{A}$
  -----------------------------------------
  $3b-2a = 5\cos{A}$
  
  $\cos{A} = \frac{3b-2a}{5}$
  
  $\sin^2{A} + \cos^2{A} = 1$
  
  $\left(\frac{3a-2b}{5}\right)^2 + \left(\frac{3b-2a}{5}\right)^2 = 1$
  • May 6th 2012, 06:14 AM
  biffboy
  Re: Trig
  In terms of answering the question you had answered it as soon as you wrote a line without A in.
  • May 6th 2012, 08:00 PM
  Prove It
  Re: Trig
  Quote:
  
  Originally Posted by biffboy
  In terms of answering the question you had answered it as soon as you wrote a line without A in.
  
  That is true, but I like to have one variable in terms of the other if possible. Just personal preference.
  
  The OP is welcome BTW ><
explanation: The document contains a math question and a detailed answer to this question. The question asks for a relation between \(a\) and \(b\) without the variable \(A\), given the equations \(a = 2\cos A + 3\sin A\) and \(b = 3\cos A + 2\sin A\). The answer provides a step-by-step derivation to express \(a\) in terms of \(b\).
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Trig
  
  • May 6th 2012, 05:06 AM
  srirahulan
  Trig
  a=2cosA+3sinA , b=3cosA+2sinA So give a relation between (a) and (b) without A.
  • May 6th 2012, 05:39 AM
  Prove It
  Re: Trig
  \displaystyle \begin{align*} a &= 2\cos{A} + 3\sin{A} \\ b &= 3\cos{A} + 2\sin{A} \\ \\ 3a &= 6\cos{A} + 9\sin{A} \\ 2b &= 6\cos{A} + 4\sin{A} \\ \\ 3a - 2b &= 6\cos{A} + 9\sin{A} - \left(6\cos{A} + 4\sin{A}\right) \\ 3a - 2b &= 5\sin{A} \\ \frac{3}{5}a - \frac{2}{5}b &= \sin{A} \\ \left(\frac{3}{5}a - \frac{2}{5}b\right)^2 &= \sin^2{A} \\ \frac{9}{25}a^2 - \frac{12}{25}ab + \frac{4}{25}b^2 &= \sin^2{A} \\ \\ 2a &= 4\cos{A} + 6\sin{A} \\ 3b &= 9\cos{A} + 6\sin{A} \\ \\ 3b - 2a &= 9\cos{A} + 6\sin{A} - \left(4\cos{A} + 6\sin{A}\right) \\ 3b - 2a &= 5\cos{A} \\ \frac{3}{5}b - \frac{2}{5}a &= \cos{A} \\ \left(\frac{3}{5}b - \frac{2}{5}a\right)^2 &= \cos^2{A} \\ \frac{9}{25}b^2 - \frac{12}{25}ab + \frac{4}{25}a^2 &= \cos^2{A} \end{align*}
  
  \displaystyle \begin{align*} \frac{9}{25}a^2 - \frac{12}{25}ab + \frac{4}{25}b^2 + \frac{9}{25}b^2 - \frac{12}{25}ab + \frac{4}{25}a^2 &= \sin^2{A} + \cos^2{A} \\ \frac{13}{25}a^2 - \frac{24}{25}ab + \frac{13}{25}b^2 &= 1 \\ a^2 - \frac{24}{13}ab + b^2 &= \frac{25}{13} \\ a^2 - \frac{24}{13}ab + \left(-\frac{12}{13}b\right)^2 - \left(-\frac{12}{13}b\right)^2 + b^2 &= \frac{25}{13} \\ \left(a - \frac{12}{13}b\right)^2 - \frac{144}{169}b^2 + \frac{169}{169}b^2 &= \frac{325}{169} \\ \left(a - \frac{12}{13}b\right)^2 + \frac{25}{169}b^2 &= \frac{325}{169} \\ \left(a - \frac{12}{13}b\right)^2 &= \frac{325 - 25b^2}{169} \\ a - \frac{12}{13}b &= \frac{\pm \sqrt{325 - 25b^2}}{13} \\ a &= \frac{12b \pm \sqrt{325 - 25b^2}}{13}\end{align*}
  • May 6th 2012, 05:54 AM
  skeeter
  Re: Trig
  Quote:
  
  Originally Posted by srirahulan
  a=2cosA+3sinA , b=3cosA+2sinA So give a relation between (a) and (b) without A.
  
  $3a = 6\cos{A}+9\sin{A}$
  $-2b = -6\cos{A}-4\sin{A}$
  -----------------------------------------
  $3a-2b = 5\sin{A}$
  
  $\sin{A} = \frac{3a-2b}{5}$
  
  $3b = 9\cos{A}+6\sin{A}$
  $-2a=-4\cos{A}-6\sin{A}$
  -----------------------------------------
  $3b-2a = 5\cos{A}$
  
  $\cos{A} = \frac{3b-2a}{5}$
  
  $\sin^2{A} + \cos^2{A} = 1$
  
  $\left(\frac{3a-2b}{5}\right)^2 + \left(\frac{3b-2a}{5}\right)^2 = 1$
  • May 6th 2012, 06:14 AM
  biffboy
  Re: Trig
  In terms of answering the question you had answered it as soon as you wrote a line without A in.
  • May 6th 2012, 08:00 PM
  Prove It
  Re: Trig
  Quote:
  
  Originally Posted by biffboy
  In terms of answering the question you had answered it as soon as you wrote a line without A in.
  
  That is true, but I like to have one variable in terms of the other if possible. Just personal preference.
  
  The OP is welcome BTW ><
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed answer to this question. The question asks for a relation between \(a\) and \(b\) without the variable \(A\), given the equations \(a = 2\cos A + 3\sin A\) and \(b = 3\cos A + 2\sin A\). The answer provides a step-by-step derivation to express \(a\) in terms of \(b\).
  
  Label: question-and-answer