Proving Identities & Solving Equations

The logic of a proof is that you do not yet know the two sides are equal — that is the thing you are trying to establish. So you may not treat the statement as an equation. You pick one side, usually the messier one, and transform it using identities you already trust until it turns into the other side. The two sides are kept in separate columns and never allowed to trade terms. The moves available to you are always the same six, in roughly this order of preference.

1. Choose the more complicated side. That is the side with room to be simplified; the simpler side is your target.
2. Convert everything on that side into sine and cosine. This is the default opening move and it works surprisingly often.
3. Look for a Pythagorean substitution. Anything of the form $1-{\sin }^2\theta$, $1-{\cos }^2\theta$, ${\sec }^2\theta -1$ or ${\csc }^2\theta -1$ is a swap waiting to happen.
4. Combine fractions over a common denominator. A sum of two fractions almost never simplifies until it is a single fraction.
5. Factor whenever you can, especially differences of squares such as ${\cos }^2\theta -{\sin }^2\theta =(\cos \theta -\sin \theta )(\cos \theta +\sin \theta )\text{.}$
6. Stop the moment your side matches the target, and write a concluding line stating that the two sides are equal.

4.5 Proving Trig IDENTITIES (full lesson) | grade 12 MHF4U | jensenmath.ca

An equation is a different animal. Here you ARE allowed to move terms across the equals sign, because the goal is to isolate the trig ratio rather than to prove anything. The standard route is to get every term onto one side, use an identity so that only ONE trig ratio appears, then factor and set each factor to zero — exactly the way you would solve a quadratic.

1. Bring everything to one side so the equation reads (expression) $=0$.
2. Use identities to get down to a single trig ratio. If you see both $\sin 2x$ and $\sin x$, expand the double angle; if you see both $\cos 2x$ and $\cos x$, choose the form of $\cos 2x$ written in cosines.
3. Factor. Treat the trig ratio as if it were a variable: $2{\sin }^{2}x+\sin x-1$ factors just like $2u^2+u-1$.
4. Set each factor equal to zero to get simple equations such as $\sin x=\frac{1}{2}$.
5. For each one, find the reference angle from the POSITIVE value, then place it in every quadrant where the sign is right.
6. Discard any solution outside the stated interval, and check nothing was lost to a division.

Your calculator's inverse trig button returns exactly one angle — the principal value — and that is where marks disappear. Over $0\le x<2\pi$ a simple sine or cosine equation normally has TWO solutions, and once you factor a quadratic you can easily end up with three or four. The fix is a habit, not a formula: find the reference angle using the positive version of the value, then use the sign of the original value together with the CAST rule to decide which quadrants to place it in. Two extra traps are worth naming here. First, if the equation involves a multiple angle such as $\sin 2x=\frac{\sqrt{3}}{2}$, then as $x$ runs over $0\le x<2\pi$ the angle $2x$ runs over $0\le 2x<4\pi$, so you must solve over the DOUBLED interval and then halve every answer — which typically yields four solutions, not two. Second, never divide both sides by a trig ratio: dividing $\sin 2x=\sin x$ by $\sin x$ silently deletes every solution where $\sin x=0$. Factor instead.

Solve a Trigonometric Equation: 2sin^2(x)=1 over the interval [0,2pi)