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acen_gr
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Is this identity possible?
[tex]cot 2x = \frac{cos3x + cosx}{sin 3x + sinx}[/tex]
Thanks!
[tex]cot 2x = \frac{cos3x + cosx}{sin 3x + sinx}[/tex]
Thanks!
Is this homework?
Is this identity possible?
[tex]cot 2x = \frac{cos3x + cosx}{sin 3x + sinx}[/tex]
Thanks!
Try cross multiplying to get:
[tex]\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)[/tex]
and then apply the half sum/difference identities to all of the sin cos products on each side of the equation. It comes out fairly easy with this step.
Use:
[tex]\sin a \cos b = \frac{1}{2} \left[ \sin(a+b) + \sin(a-b) \right][/tex]
[tex]\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)[/tex]
[tex]\cos 2x \sin3x + \cos2x sinx = \sin2x \cos3x + \sin2x cosx[/tex]
Should I end here? I think both doesn't equal up. Or should I go further by extracting cos2x, sin2x, cos3x, and sin3x ?
No, apply the half sum-difference formula (that I gave above) to each of the sin-cos products.[tex]\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)[/tex]
[tex]\cos 2x \sin3x + \cos2x sinx = \sin2x \cos3x + \sin2x cosx[/tex]
Should I end here?