Solve for #cos(x)#, when #sin(x)=2/3# and #0^@<x<90^@#?

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Answer 1 · 2017-11-20T02:58:46

#cos(x) = sqrt5/3#

Given:

#sin(x)= 2/3; 0^@ < x < 90^@#

Use the identity:

#cos(x) = +-sqrt(1 - sin^2(x))#

Because we are given that #0^@ < x < 90^@# we know that we must select the positive value:

#cos(x) = sqrt(1 - sin^2(x))#

Substitute #(2/3)^2# for #sin^2(x)#

#cos(x) = sqrt(1 - (2/3)^2)#

#cos(x) = sqrt(9/9 - 4/9)#

#cos(x) = sqrt(5/9)#

#cos(x) = sqrt5/3#

Answer 2 · 2017-11-20T03:19:42

#cos(x)=sqrt5/3#

#0^@ < x < 90^@# implies that the angle is in the #"1"^"st"# quadrant.

#sin(x)=2/3#
enter image source here

Using Pythagoras' theorem,

#a=sqrt(3^2-2^2)#
#color(white)(a)=sqrt5#

#cos(x)="adjacent"/"hypotenuse"#
#color(white)(cos(x))=sqrt5/3#