basic trig question (1 Viewer)

Sien · Jun 12, 2015

If sinx= 2/3 and cosy=3/4 and x and y are acute angles, find the exact value of a. cosx b. siny
I just keep getting weird answers, so it would be nice if you could post working out thanks

kawaiipotato · Jun 12, 2015

If x and y are acute angles, it must lie in the first quadrant, hence sinx, siny, cosx, cosy must be positive
$$ Given $ sinx = \frac{2}{3}$
$Draw up a right-angled triangle, with an angle of 'x' and opposite side being 2 and hypotenuse being 3.$
$the sine of this angle equals 2/3 and satisfies the given condition$
$Find the unknown side by pythagoras theorem, adjacent side = \sqrt{3^2 - 2^2} = \sqrt{5}$
$\therefore cosx = \frac{\sqrt{5}}{3}$

$alternate method$
$sinx = \frac{2}{3}$
$sin^{2}x = \frac{4}{9}$
$1 - cos^{2}x = \frac{4}{9}$
$\frac{5}{9} = cos^{2}x$
$cosx = \pm \frac{\sqrt{5}}{3}$
$$ since x is acute, $ cosx > 0$
$cosx = \frac{\sqrt{5}}{3}$

calebwhitaker199 · Jun 12, 2015

Hope that helps

Sien · Jun 12, 2015

calebwhitaker199 said:
Hope that helps

kawaiipotato said:
If x and y are acute angles, it must lie in the first quadrant, hence sinx, siny, cosx, cosy must be positive
$$ Given $ sinx = \frac{2}{3}$
$Draw up a right-angled triangle, with an angle of 'x' and opposite side being 2 and hypotenuse being 3.$
$the sine of this angle equals 2/3 and satisfies the given condition$
$Find the unknown side by pythagoras theorem, adjacent side = \sqrt{3^2 - 2^2} = \sqrt{5}$
$\therefore cosx = \frac{\sqrt{5}}{3}$

$alternate method$
$sinx = \frac{2}{3}$
$sin^{2}x = \frac{4}{9}$
$1 - cos^{2}x = \frac{4}{9}$
$\frac{5}{9} = cos^{2}x$
$cosx = \pm \frac{\sqrt{5}}{3}$
$$ since x is acute, $ cosx > 0$
$cosx = \frac{\sqrt{5}}{3}$

thanks

Speed6 · Jun 12, 2015

Did the same question today

basic trig question (1 Viewer)

Sien

将来: NEET

kawaiipotato

Well-Known Member

calebwhitaker199

New Member

Sien

将来: NEET

Speed6

Retired '16

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