## Trigonometry, Trigonometric formulas, IB Math HL

Discussions for the Core part of the syllabus. Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths HL Revision Notes

### Trigonometry, Trigonometric formulas, IB Math HL

IB Mathematics HL – Trigonometry, Trigonometric identities, double angle formula

How can we simplify the following trigonometric expression?

$\frac{sin4x(1-cos2x)}{cos2x(1-cos4x)}$

Thanks
nicole

Posts: 0
Joined: Mon Jan 28, 2013 8:10 pm

### Re: Trigonometry, Trigonometric formulas, IB Math HL

IB Math HL – Trigonometry, Trigonometric identities, double angle formula

$\frac{sin4x(1-cos2x)}{cos2x(1-cos4x)}$

We’ll use the following double angle formulas:

$sin2x=2sinxcosx$ and $sin^2x=\frac{1}{2}(1-cos2x)$

So, the expression can be written as

$\frac{sin4x(1-cos2x)}{cos2x(1-cos4x)}=$

$=\frac{2sin2xcos2x \cdot 2sin^2x }{cos2x \cdot 2sin^2(2x) }=$

$=\frac{2sin^2x }{sin(2x) }=\frac{2sin^2x }{2sinxcosx }=$

$=\frac{sinx }{cosx }=tanx$

Hope these help!!
miranda

Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm

### Re: Trigonometry, Trigonometric formulas, IB Math HL

Thank you!!
nicole

Posts: 0
Joined: Mon Jan 28, 2013 8:10 pm