How do you use double angle formulas to calculate cos 2x and sin 2x without finding x if #cos x = 3/5# and x is in the first quadrant?

2 Answers
Jun 25, 2018

#cos(2x)=-7/25#
#sin(2x)=24/25#

Explanation:

The double-angle formulas state that:
#cos(2x)=cos^2(x)-sin^2(x)#

#sin(2x)=2sin(x)cos(x)#

Now, we are given that #cos(x)=3/5#. Then, knowing that #sin^2(x)+cos^2(x)=1# and that #x# is in the first quadrant (and thus #sin(x)# and #cos(x)# are both positive), we can find that #sin(x)=sqrt(1-cos^2(x))=4/5#.

Thus, using the identities mentioned above,
#cos(2x)=cos^2(x)-sin^2(x)=(3/5)^2-(4/5)^2=-7/25#

#sin(2x)=2sin(x)cos(x)=2*4/5*3/5=24/25#

Jun 25, 2018

# cos2x = -7/25 #
# sin2x = 24/25 #

Explanation:

We have #cosx=3/5# and #x# is acute.

Using #sin^2x + cos^2x -= 1 #, we have:

#sin^2x + (3/5)^2 = 1 #
# => sin^2x = 1-9/25 #
# :. \ sin^2x = 16/25 #
# :. \ \ sinx = +-sqrt(16/25) #

And knowing that #x# is acute we take the positive solution.

# => sinx = 4/5 #

Next we use the sine and cosine double angle formula, so that:

# cos2x -= cos^2x - sin^2 x #

# \ \ \ \ \ \ \ \ \ = (3/5)^2 - (4/5)^2 #

# \ \ \ \ \ \ \ \ \ = 9/25 - 16/25 #

# \ \ \ \ \ \ \ \ \ = -7/25 #

And:

# sin2x -= 2sinxcosx #

# \ \ \ \ \ \ \ \ \ = 2(3/5)(4/5) #

# \ \ \ \ \ \ \ \ \ = 24/25 #