If -180o <= 180o , solve the equation
cos(2theta - 70o) = sin(theta+40o)
A good start would probably to apply sum and difference identities, so you can get $\displaystyle \sin(\theta)$ and $\displaystyle \cos(\theta)$ by themselves, and then work from there.
. . . . .$\displaystyle \cos(2\theta\, -\, 70^{\circ})\, =\, \cos(2\theta)\cos(70^{\circ})\, +\, \sin(2\theta)\sin(70^{\circ})$
. . . . .$\displaystyle \sin(\theta\, +\, 40^{\circ})\, =\, \sin(\theta)\cos(40^{\circ})\, +\, \cos(\theta)\sin(40^{\circ})$
And note also the double-angle identities:
. . . . .$\displaystyle \cos(2\theta)\, =\, \cos^2(\theta)\, -\, \sin^2(\theta)$
. . . . .$\displaystyle \sin(2\theta)\, =\, 2\sin(\theta)\cos(\theta)$
See where this leads....
If you get stuck please reply showing your steps and how far you have gotten. Thank you!
Type [tex]\theta[/tex].
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