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Answer 1 · 2017-12-13T06:48:57

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

We cant express this in terms of #x# but we can simplify the expression

One thing that we can consider is how #cos(2theta)= cos(theta + theta ) #

Now we can use one of our additional formulae:

#cos(A+B) = cosAcosB - sinAsinB#

Applying this formula for #cos(2theta):#

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

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

We can also simplify this further:

Use our trig identities:

# (1-sin^2 theta ) - sin theta #

#=> 1 - 2sin^2 theta #

Or...

# cos^2 theta - (1 - cos^2 theta ) #

#=> 2cos^2 theta - 1 #

These are particulaly useful for integration!