If #sintheta=1/3# and #theta# is in quadrant I, how do you evaluate #cos2theta#?

2 Answers
Jun 5, 2018

Answer:

#7/9#

Explanation:

#cos(2theta) #

= #cos^2theta-sin^2theta#

=#1-sin^2theta-sin^2theta#

=# 1-2sin^2theta#

=#1-2(1/3)^2#

=#1-2/9#

=#7/9#

Jun 5, 2018

Answer:

#cos2theta=7/9#

Explanation:

#"using the "color(blue)"trigonometric identities"#

#•color(white)(x)cos2theta=cos^2theta-sin^2theta#

#•color(white)(x)sin^2theta+cos^2theta=1#

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

#theta" is in first quadrant as is "2theta#

#costheta=sqrt(1-(1/3)^2)=sqrt(1-1/9)=sqrt(8/9)#

#rArrcostheta=(2sqrt2)/3#

#cos2theta=((2sqrt2)/3)^2-(1/3)^2#

#color(white)(cos2theta)=8/9-1/9=7/9#