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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?

Answer 1

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

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#

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Answer 2

James Adkins

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

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 #

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Answer 3

Zoe Austin

To calculate (\cos 2x) and (\sin 2x) without finding (x), if (\cos x = \frac{3}{5}) and (x) is in the first quadrant, use the double angle formulas:

[ \cos 2x = 2(\cos^2 x) - 1 ] [ \sin 2x = 2\sin x \cos x ]

Given (\cos x = \frac{3}{5}), first find (\sin x) using the Pythagorean identity:

[ \sin^2 x + \cos^2 x = 1 ] [ \sin^2 x = 1 - \cos^2 x ] [ \sin^2 x = 1 - \left(\frac{3}{5}\right)^2 ] [ \sin^2 x = 1 - \frac{9}{25} ] [ \sin^2 x = \frac{25 - 9}{25} ] [ \sin^2 x = \frac{16}{25} ] [ \sin x = \pm \frac{4}{5} ]

Since (x) is in the first quadrant, (\sin x = \frac{4}{5}).

Now, use the double angle formulas:

[ \cos 2x = 2\left(\frac{3}{5}\right)^2 - 1 ] [ \cos 2x = 2\left(\frac{9}{25}\right) - 1 ] [ \cos 2x = \frac{18}{25} - 1 ] [ \cos 2x = \frac{18}{25} - \frac{25}{25} ] [ \cos 2x = -\frac{7}{25} ]

[ \sin 2x = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} ] [ \sin 2x = \frac{24}{25} ]

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