Trigonometry

How does sin^2(2x)= (1-cos(4x))/2 ? I just can't work it out!

Thanks!

How does sin^2(2x)= (1-cos(4x))/2 ? I just can't work it out!

Thanks!

Re-arrange the double angle formulae for

\cos 2A

to get:

\displaystyle \cos 2A = 1 - 2\sin^2 A \iff \sin^2 A = \frac{1- \cos 2A}{2}

, agree?

Then set

A = 2x

.

Also, moved to maths.

Re-arrange the double angle formulae for

\cos 2A

to get:

\displaystyle \cos 2A = 1 - 2\sin^2 A \iff \sin^2 A = \frac{1- \cos 2A}{2}

, agree?

Then set

A = 2x

.

Also, moved to maths.

Thank you so much! It seems so simple now! I was trying to figure it out for ages!

Thank you so much! It seems so simple now! I was trying to figure it out for ages!

No problem, it comes in useful all the time when integrating squared trigonometric functions. Remember that also:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\cos 2A = 2 \cos^2 A - 1 \iff \cos^2 A = \frac{\cos 2A + 1}{2}\end{equation*}

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