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

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

Postby bazbeatz on Sun Jun 21, 2009 4:53 pm

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
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Postby stapel_eliz on Sun Jun 21, 2009 7:16 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:

. . . . .

Squaring is a good idea, I think:

. . . . .

. . . . .

. . . . .

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

. . . . .

. . . . .

Can you take it from there? :wink:
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Re: trig: solving 4sin'theta'cos'theta' + 1 = 2(sin'theta' + cos

Postby bazbeatz on Sun Jun 21, 2009 10:23 pm

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
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Postby stapel_eliz on Sun Jun 21, 2009 11:53 pm

Check the math:

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

:wink:
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