If sinθ =4/7, what is cosθ?

3 Answers
Jun 20, 2018

Answer:

#cosθ= +- 33^(1/2)/7#

Explanation:

Since

#Sin^2θ xx cos^2θ =1 #

you have

#(4/7)^2 xx cos^2θ=1#

So

#Cos^2θ= 1- 16/49#

#Cosθ= +- 33^(1/2)/7 #

Jun 20, 2018

Answer:

#cos(theta)=+-sqrt(33)/7#

Explanation:

We want to find #cos(theta)#, when #sin(theta)=4/7#

By the pythagorean trig identity

#color(blue)(cos^2(theta)+sin^2(theta)=1#

#=>cos^2(theta)=1-sin^2(theta)#

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

Substitute #color(red)(sin(theta)=4/7#

#cos(theta)=+-sqrt(1-(4/7)^2)#

#color(white)(cos(theta))=+-sqrt(1-(16/49)#

#color(white)(cos(theta))=+-sqrt((49-16)/49)#

#color(white)(cos(theta))=+-sqrt(33)/7#

Jun 20, 2018

Answer:

#cos(theta)=pmsqrt(33)/7#

Explanation:

Using the relation
#sin^2(theta)+cos^2(theta)=1# then we get
#cos(theta)=pmsqrt(1-16/49)#