Trigonometry, Trigonometric formulas, IB Math HL

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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