$$\cos(\theta-60^\circ)=\frac{1}{2}\sin(\theta)$$ $$2\cos(\theta - 60^\circ)=\sin(\theta)$$ $$2=\frac{\sin(\theta)}{\cos(\theta-60^\circ)}$$

How do I get rid of the $60^\circ$ so that I can make this a solvable trig identity?

Edit: This is my whole solution so you can see where I am going wrong: $$\cos(\theta-60^\circ)=\frac{1}{2}\sin(\theta)$$ $$2\cos(\theta)\cos(60^\circ)+2\sin(\theta)\sin(60^\circ)=\sin(\theta)$$ $$\frac{2\cos(\theta)+2\sin(\theta)(\sqrt3)}{2}=\sin(\theta)$$ $$\cos(\theta)+\sin(\theta)(\sqrt3)=\sin(\theta)$$ $$\frac{\cos(\theta)}{\sin(\theta)}+\sqrt3=\frac{\sin(\theta)}{\sin(\theta)}$$ $$\cot(\theta)+\sqrt3=1$$ $$\cot(\theta)=1-\sqrt3$$ $$\theta=\cot^{-1}(1-\sqrt3)$$

Thanks in advance

  • 1
    $\begingroup$ Hint: think trig identities: cos(u - v) = cos u cos v + sun u sin v $\endgroup$ – Amzoti Nov 13 '12 at 16:03
  • $\begingroup$ Oh well, it seems I am right lol... thanks for the help anyway guys $\endgroup$ – user866190 Nov 13 '12 at 16:41

In general, $$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin( b).$$

In your case, we have $$\cos(\theta - 60^\circ) = \cos(60^\circ - \theta) =\cos(60^\circ) \cos\theta+\sin(60^\circ) \sin\theta.$$

You can substitute this into your first line/equation. Then resulting equation will be: $$\cos(\theta-60^\circ)=\frac{1}{2}\sin(\theta)$$ $$\cos(60^\circ) \cos\theta+\sin(60^\circ) \sin\theta=\frac{1}{2}\sin\theta$$ $$ 2\cos(60^\circ) \cos\theta+ 2\sin(60^\circ) \sin\theta=\sin\theta.$$

You can then substitute the values of $\sin(60^\circ)$ and $\cos(60^\circ)$ into the equation, and then divide both sides by $\sin\theta$.

The rests is "simply" simplification.

UPDATE: Regarding your edited work: I think you did just fine!

  • $\begingroup$ @BabakSorouh I was close to your answer! $\endgroup$ – user866190 Nov 13 '12 at 16:24

I think the answer given in the posted question is correct. Why would it be asserted that it is not?

(Of course want all of the "multiple values" of the inverse cotangent function at the given point would be solutions.)


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