Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

Question #86887

1 Answer

Oct 27, 2017

$2 cos (2 x)$ $2cos(2x)$ or $2 - 4 {sin}^{2} (x))$ $2-4sin^2(x))$

Explanation:

$sin (2 a) = 2 sin (a) cos (a)$ $sin(2a)=2sin(a)cos(a)$
$cos (2 a) = 1 - 2 {sin}^{2} (a)$ $cos(2a)=1-2sin^2(a)$

therefore:

$\frac{sin (4 x)}{sin (2 x)} = \frac{2 sin (2 x) cos (2 x)}{sin (2 x)} = 2 cos (2 x)$ $sin(4x)/sin(2x)=(2sin(2x)cos(2x))/sin(2x)=2cos(2x)$

$= 2 (1 - 2 {sin}^{2} (x)) = 2 - 4 {sin}^{2} (x))$ $=2(1-2sin^2(x))=2-4sin^2(x))$

Related questions

Impact of this question

1128 views around the world

You can reuse this answer
Creative Commons License