# finding.the.value.of a trigonometric ratio

• Nov 25th 2012, 01:20 AM
Tutu
finding.the.value.of a trigonometric ratio
Hi I am wondering about my answer..Given that cosA-sinA=1/3,find the exact value of cosAsinA. I got (2sqrt(5))/9 but the answer is 4/9.. Really appreciate this thanksa. Lot!!
• Nov 25th 2012, 01:28 AM
MarkFL
Re: finding.the.value.of a trigonometric ratio
Take the given relation and square it:

$\displaystyle (\cos(A)-\sin(A))^2=\left(\frac{1}{3} \right)^2$

You should then find the given result, after you simplify and solve for the required expression.
• Nov 25th 2012, 01:30 AM
Prove It
Re: finding.the.value.of a trigonometric ratio
Quote:

Originally Posted by Tutu
Hi I am wondering about my answer..Given that cosA-sinA=1/3,find the exact value of cosAsinA. I got (2sqrt(5))/9 but the answer is 4/9.. Really appreciate this thanksa. Lot!!

\displaystyle \displaystyle \begin{align*} \cos{(A)} - \sin{(A)} &= \frac{1}{3} \\ \left[ \cos{(A)} - \sin{(A)} \right]^2 &= \left( \frac{1}{3} \right)^2 \\ \cos^2{(A)} - 2\cos{(A)}\sin{(A)} + \sin^2{(A)} &= \frac{1}{9} \\ 1 - 2\cos{(A)}\sin{(A)} &= \frac{1}{9} \\ 1 - \frac{1}{9} &= 2\cos{(A)}\sin{(A)} \\ \frac{8}{9} &= 2\cos{(A)}\sin{(A)} \\ \frac{4}{9} &= \cos{(A)}\sin{(A)} \end{align*}