Double formula for sin/cos?

Question

Either this is a mistake, or I memorized the formula wrong. My answer key is saying to use cos(2a)=2sin(a)cos(a) to solve a problem. Isn't that the formula for double sine?

Answer 1

Go to the link below.  There is a lot of junk there, but what you are looking for is under 6.1 in the contents box at the top.

Answer 2

You are correct. It should say sin(2a)=2sin(a)cos(a)

Answer 3

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

cos(2x)

= 2cos(x)^2-1

=1- 2 sin(x)^2

= cos(x)^2-sin(x)^2

If you can remember the rule for adding angles in sine and cosine, you should be able to derive all these formulae. For example;

sin (2x)

= sin (x + x)

= sin(x) cos(x) + cos(x) sin(x)

= 2 sin(x) cos(x)

cos (2x)

= cos (x + x)

= cos (x) cos(x) - sin(x) sin(x)

=cos(x)^2 - sin(x)^2

Since we also know that sin(x)^2 + cos(x)^2 =1

cos (2x)

=cos(x)^2 - sin(x)^2

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

=2cos(x)^2-1

OR

cos (2x)

=cos(x)^2 - sin(x)^2

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

=1- 2 sin(x)^2

Rather than memorizing a whole heap of formula's, it is often much easier to remember the most basic ones. The two basic ones I remember are the addition rule and the basic Pythagorean identity; armed with these two, you can basically derive most of the other important trigonometric identities.