Question #356fc

Recall that #cos^2x +sin^2x= 1#.

For #x#

#(3/5)^2 + cos^2x = 1#

#cos^2x= 16/25#

#cosx = +-4/5#

Since we're in the first quadrant, cosine must be positive, thus #cosx = 4/5#.

For #y#

#(7/25)^2 + sin^2y = 1#

#sin^2y = 1 - 49/625#

#siny = +-24/25#

Once again, as we're in the first quadrant, both sine and cosine are positive, thus #siny = 24/25#.

Hopefully this helps!

Pythagorean triples of 3,4,5 and 7,24, 25. And nothing is negative because this is all in quadrant one. I suggest that you draw the triangles and then solve for the third unknown side with Pythagorean theorem if you would like. And once you have the sides you need for both the triangles, just remember SOH CAH TOA.
Sin=Opposite/Hypotenuse
Cosine=Adjacent/Hypotenuse
Tan= Opposite/Adjacent

Use the identity:

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

Substitute #sin^2(x) = (3/5)^2#

#cos(x)=+-sqrt(1-(3/5)^2)#

Because are given that x is in the first quadrant, we know to use the positive value:

#cos(x)=sqrt(1-(3/5)^2)#

Simplify:

#cos(x)=sqrt(25/25-9/25)#

#cos(x)=sqrt(16/25)#

#cos(x)=4/5#

Use the identity:

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

Substitute #cos^2(y) = (7/25)^2#

#sin(y) = +-sqrt(1-(7/25)^2)#

Because are given that y is in the first quadrant, we know to use the positive value:

#sin(y) = sqrt(1-(7/25)^2)#

Simplify:

#sin(y) = sqrt(625/625-49/625)#

#sin(y) = sqrt(576/625)#

#sin(y) = 24/25#