trig: solving 4sin'theta'cos'theta' + 1 = 2(sin'theta' + cos

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trig: solving 4sin'theta'cos'theta' + 1 = 2(sin'theta' + cos

Ok this may be more of an algebra question than a trig one but it originated from a trig one. Any help would be much appreciated...

Solve the following equation for theta giving your answers in the interval from -180 to 180:

4sin'theta'cos'theta' + 1 = 2(sin'theta' + cos'theta')

After multiplying out the brackets, squaring the expression and substituting cos^2'theta' for 1-sin^'theta' I get the following:

16sin^2'theta' - 16sin^4'theta' = 3

However I can't seem to simplify this further to enable me to find 'theta'. Can someone show me how to do this...and say if the approach I have used is the quickest way of solving this equation?

Thanks
bazbeatz

Posts: 4
Joined: Sun Jun 07, 2009 5:48 pm

bazbeatz wrote:Solve the following equation for theta giving your answers in the interval from -180 to 180:

4sin'theta'cos'theta' + 1 = 2(sin'theta' + cos'theta')

You have the following:

. . . . .$4\sin(\theta) \cos(\theta)\, +\, 1\, =\, 2(\sin(\theta)\, +\, \cos(\theta))$

Squaring is a good idea, I think:

. . . . .$16\sin^2(\theta)\cos^2(\theta)\, +\, 8\sin(\theta)\cos(\theta)\, +\, 1\, =\, 4\sin^2(\theta)\, +\, 8\sin(\theta)\cos(\theta)\, +\, 4\cos^2(\theta)$

. . . . .$16\sin^2(\theta)\cos^2(\theta)\, +\,1\, =\, 4$

. . . . .$16\sin^2(\theta)\cos^2(\theta)\, =\, 3$

But now use sin(2x) = 2sin(x)cos(x) to get:

. . . . .$(2\sin(2\theta))^2\, =\, 3$

. . . . .$\sin(2\theta)\, =\, \pm \frac{\sqrt{3}}{2}$

Can you take it from there?

stapel_eliz

Posts: 1705
Joined: Mon Dec 08, 2008 4:22 pm

Re: trig: solving 4sin'theta'cos'theta' + 1 = 2(sin'theta' + cos

Hi,

Thanks for that. I can now get the correct answer. However I'm just puzzled by one step in your calcs (which is probably where I went wrong in the first place)...

When evaluating (4 sin 'theta' cos 'theta')^2 you give the answer containing the term 8sin'theta'cos'theta'. However I would have thought the answer is simply 16sin^2'theta'cos^2'theta' in the same way as (abc)^2 = a^2 b^2 c^2. Am I missing something here?

Thanks
bazbeatz

Posts: 4
Joined: Sun Jun 07, 2009 5:48 pm

Check the math:

(4x + 1)(4x + 1) = 16x^2 + 4x + 4x + 1 = 16x^2 + 8x + 1

stapel_eliz

Posts: 1705
Joined: Mon Dec 08, 2008 4:22 pm