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# i need help with trigonometry please

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

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sinx = opposite/ hypotenuse

so when sin x =1/3

opposite side of a right angled triangle = 1

hypotenuse 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

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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)] )

= [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