How do you work with trigonometric functions?

Question

How can you solve for something such as cos^2 2x - sin ^2 2x = 0 using trigonometric functions?

Answer 1

This can be easily solved as follows:
cos² 2x - sin² 2x = 0
Thus cos² 2x = sin² 2x
Hence sin² 2x/cos² 2x = 1
ie tan² 2x = 1
Thus tan 2x = +1 or - 1
thus 2x = pi/4 + kpi/2 where k is any integer
whence x = pi/8 + kpi/4

Answer 2

You have to use your trigonometric identities. In this case, try:

cos^2( z ) + sin^2( z ) = 1

for any z. In other words,

sin^2(z) = 1 - cos^2(z)

If you set z=2x, then you can substitute this into your equation. You get:

cos^2 2x - (1 - cos^2 2x) = 0

which is:

cos^2 2x - 1 + cos^2 2x = 0

which is:

2*cos^2 2x = 1

and

cos^2 2x = 1/2

Now, you can take the square root of both sides and get:

cos 2x = +/- 1/sqrt(2)

Now you have to figure out every angle for which cos(angle)=1/sqrt(2) and every angle for which cos(angle)=-1/sqrt(2). In other words, you need to solve:

2x = arccos( 1/sqrt(2) )

and

2x = arccos( -1/sqrt(2) )

cos(angle) is equal to 1/sqrt(2) at:

pi/4 + 2*k*pi

where k is any integer. cos(angle) is equal to -1/sqrt(2) at:

3*pi/4 + 2*k*pi

where k is any integer.

In other words, cos(angle) is either 1/sqrt(2) or -1/sqrt(2) at every angle of the form:

pi/4 + k*pi

where k is any integer.

Now, to solve for x, you need to divide these angles by 2. So your final answer is:

x = pi/8 + k*pi/2

where k is any integer.

Just test it out:

cos^2( 2*pi/8 ) - sin^2( 2*pi/8 ) = 1/2 - 1/2 = 0

Answer 3

cos^2 2x - sin^2 2x = 0
cos^2 2x + xin^2 2x = 1

Lets sum:

2 cos^2 2x = 1

cos^2 2x = 1/2

cos 2x = 1/sqrt 2

So 2x = pi / 4 + 2kpi, k an integer
or
2x= 3 pi/4 + 2k pi

And x = pi/8 + k pi or x = 3pi/8 + k pi

Ana

Answer 4

Notice that this is a difference of squares.
a^2 - b^2 = (a-b)(a+b)
(cos 2x)^2 - (sin 2x)^2 = 0 becomes
(cos(2x) - sin(2x))(cos(2x) + sin(2x)) = 0
Set each part equal to zero.
cos(2x) - sin(2x) = 0 or cos(2x) + sin(2x) = 0
Now solve for x.