Posted July 6, 2014 hello, i'm stuck with a trigonometry question! given that sinx=1/3 , where x is an acute angle, find the exact value of(a) cosx(b) cos2x doen anyone have the mark scheme for this? or do you know how to solve it?thank you very much Share this post Link to post Share on other sites

Posted July 6, 2014 sinx = opposite/ hypotenuse so when sin x =1/3 opposite side of a right angled triangle = 1hypotenuse of a right angled triangle = 3 using pythag, the adjacent side of the triangle = squroot(9-1) = sqroot(8) a) cos x = adjacent/hypotenuse = sqroot(8)/3 b) cos 2x = cos^2(x) - sin^2(x) = 2 cos^2(x) - 1 = 1 - 2 sin^2(x) so sub in cos/ sin into one of the forms of cos2x Let's take cos2x = 1 - 2sin^2(x) = 1 - 2(1/3)^2= 1- 2/9= 7/9 1 person likes this Share this post Link to post Share on other sites

Posted July 13, 2014 Since x is acute, it is found in the 3rd quadrant. All trigonometric values are thus positive. Draw a right-angled triangle with opposite = 1 and hypothenuse = 3 (since sin = opposite/hypothenuse) From this triangle, use Pythagorus to solve the last side. (which is equal to [sqrt(8)] ) cos x = adjacent/hypothenuse = [sqrt(8)] / 3 To find cos 2x use the double angle identity 1 - 2sin^2 (x) (you can use another identity but this one is easier since you already have to value of sin x This can be re-written as cos 2x = 1 - 2(sin(x))^2 = 1 - 2(1/3)^2 = 1 - 2 (1/9) = 1- (2/9) = 7/9 VoilÃ Share this post Link to post Share on other sites