How do you find sinØ if cos2Ø=3/5, and Ø terminates in Quadrant 1?

3 Answers
Hriman
Mar 7, 2018

#sintheta= sqrt5/5#

Explanation:

Alright, here we know that both sin and cos are positive because this is in the first quadrant.
We will utilize the cos double angle identity to solve for #theta#
#(cos2theta)=3/5#
#(cos^2theta-sin^2theta)=3/5#
#(cos^2theta-(1-cos^2theta))=3/5#
#(2cos^2theta-1)=3/5#
#2cos^2theta =3/5+5/5#
#cos^2theta =8/5*1/2#
#cos^2theta =4/5#

#costheta =2/sqrt5#
We can solve for theta if we really want by arccos, since the value is not on the unit circle, but no point since the question is asking for the value of sin theta

From that value we know that one of the legs of the triangle is 2, and the hypotenuse is 5, using the pythagorean theorem becomes essential to find the opposite side
#B= sqrt(C^2-A^2)#
#B= sqrt((sqrt5)^2-(2)^2)#
#B= +-1#
Since sin and cos are positive, we will use positive one as the opposite side, therefore #sintheta=1/sqrt5 or sqrt5/5#

Steve M
Mar 7, 2018

# sin phi = 1/sqrt(5) #

Explanation:

We can use the identity:

# cos 2 phi -= cos^2 phi - sin^2 phi #
# \ \ \ \ \ \ \ \ \ -= (1-sin^2 phi) - sin^2 phi #
# \ \ \ \ \ \ \ \ \ -= 1-2sin^2 phi #

So we can write:

# 3/5 = 1-2sin^2 phi #
# :. 2sin^2 phi = 1-3/5 #
# :. 2sin^2 phi = 2/5 #
# :. sin^2 phi = 1/5 #

# :. sin phi = +- 1/sqrt(5) #

But we know that #phi # is in quadrant 1, so #phi# is acute, and #sin phi# is positive, thus we discard the negative solution and deduce that:

# sin phi = 1/sqrt(5) #

Aaron S.
Mar 7, 2018

#sintheta=1/sqrt(5)#

Explanation:

Trig Identities

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

#cos(2theta)=2costheta-1# #(2)#

now to the working out

#cos(2theta)=3/5#

so using #(2)#

#3/5=2cos^2theta-1#

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

add 1 and #3/5#

#8/5*1/2=(cancel2cos^2theta)/cancel2#

#8/10=cos^2theta# #(3)#

substitute #(3)# into #(1)#

#sin^2thetacancel(+8/10-8/10)=1-8/10#

#sin^2theta=10/10-8/10#

#sin^2theta=2/10#

#sin^2theta=1/5#

#cancelsqrt(sin^cancel(2)theta)=sqrt(1/5)#

#sintheta=1/(sqrt(5))#

If you want to rationalise the denominator

#sintheta=1/sqrt(5)*sqrt(5)/sqrt(5)#

#sintheta=sqrt(5)/5#

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