1. ## Trig

a=2cosA+3sinA , b=3cosA+2sinA So give a relation between (a) and (b) without A.

2. ## Re: Trig

\displaystyle \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 \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*}

3. ## Re: Trig

Originally Posted by srirahulan
a=2cosA+3sinA , b=3cosA+2sinA So give a relation between (a) and (b) without A.
$\displaystyle 3a = 6\cos{A}+9\sin{A}$
$\displaystyle -2b = -6\cos{A}-4\sin{A}$
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$\displaystyle 3a-2b = 5\sin{A}$

$\displaystyle \sin{A} = \frac{3a-2b}{5}$

$\displaystyle 3b = 9\cos{A}+6\sin{A}$
$\displaystyle -2a=-4\cos{A}-6\sin{A}$
-----------------------------------------
$\displaystyle 3b-2a = 5\cos{A}$

$\displaystyle \cos{A} = \frac{3b-2a}{5}$

$\displaystyle \sin^2{A} + \cos^2{A} = 1$

$\displaystyle \left(\frac{3a-2b}{5}\right)^2 + \left(\frac{3b-2a}{5}\right)^2 = 1$

4. ## Re: Trig

In terms of answering the question you had answered it as soon as you wrote a line without A in.

5. ## Re: Trig

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 ><