This problem is from S.L. Loney's book on Plane Trigonometry. To solve the following equation:

$$\cos\theta – \sin3\theta = \cos2\theta$$

Now, my initial approach was to algebraically manipulate the equation to yield a difference of cosines:

$$\cos\theta– \cos2\theta = \sin3\theta$$

From this, the sum formula could be applied to the RHS to yield

$$\cos\theta – \cos2\theta = \sin2\theta \cos\theta + \cos2\theta \sin\theta$$

Now, this equation is true if $\sin2\theta = 1$ and $\sin\theta = –1$. Hence $\theta$ must be equal to $1\over2$$(n\pi + (–1)^n$$\pi\over2$$)$ and equal to $n\pi – (–1)^n$$\pi\over2$

For one, I do not know how to combine the two equations into one form, or if it is even needed to. Secondly, I have been informed that there are even more values that satisfy the equation. How am I to find them?

I also don't know if the solution I provided is even accurate, so could the reader also perhaps check if the reasoning I provided is correct?

I am still relatively new in the field of trigonometry, hence your time would be much appreciated. Thank you in advance.

  • 2
    $\begingroup$ If $\sin\theta=-1$, then $\theta=-\frac\pi2+2n\pi$, for some $n\in\Bbb Z$. But then $2\theta=-\pi+4n\pi$, and therefore $\sin2\theta=0\ne1$. $\endgroup$ Jan 5, 2023 at 20:05

1 Answer 1


HINT…starting with $$\cos\theta-\cos2\theta=\sin3\theta,$$

Apply to the LHS the formula $$\cos A-\cos B=-2\sin\left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$$

Apply to the RHS the formula $$\sin2A=2\sin A\cos A$$

You can then factorise the whole equation and solve.

  • $\begingroup$ Thank you for the helpful answer! Yet, another question occurs to me: how do I know how many different solutions exist for one equation? I might solve for all solutions but one; Is there a specific pool of answers that exists for trigonometric equations? So far I've found that most equations have 2 or 3 different pools of solutions. Also, is the reasoning I provided in my post incorrect? Thank you for your time. $\endgroup$
    – Camelot823
    Jan 6, 2023 at 5:15
  • $\begingroup$ As far as I know there are no hard and fast rules for how many “pools” of solutions may exist for a given trig equation. As far as your workings are concerned, it has already been pointed out by Jose Carlos Santos that your method does not provide valid solutions because the two statements $\sin 2\theta=1$ and $\sin\theta=-1$ are inconsistent with each other. $\endgroup$ Jan 6, 2023 at 9:43
  • $\begingroup$ Also, using your method, I don't see how $sin2A = 2sinAcosA$ can be applied to the RHS; since the RHS contains $sin3A$. What was your reasoning when stating this? And if there are no rules for finding the quanitity of pools of solutions, how do you know what you're looking for? Or if any other solutions exist, and when to stop searching? $\endgroup$
    – Camelot823
    Jan 6, 2023 at 14:11
  • $\begingroup$ $\sin3A=2\sin\frac{3A}{2}\cos\frac{3A}{2}$ and both sides will contain $\sin\frac{3A}{2}$ as a common factor. $\endgroup$ Jan 6, 2023 at 14:37
  • $\begingroup$ Oh my, I haven't connected that. Thank you greatly for the help! $\endgroup$
    – Camelot823
    Jan 6, 2023 at 17:04

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