What "tricks" are there that could help verify trigonometric identities?
For example one is:
$$a\cos\theta+b\sin\theta = \sqrt{a^2+b^2}\,\cos(\theta-\phi)$$
Trigonometry – Useful Tricks for Verifying Trigonometric Identities
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In this context, I would recommend either:
- don't use $\equiv$ at all, or
- only use $\equiv$ to indicate implicit universal quantification.
Let me explain what I mean by (2).
There's this concept of universal quantification that all mathematics is based on. Its really important. If you understand it, math probably makes sense to you. If you don't, it probably doesn't. Unfortunately, this concept usually isn't taught until very late into one's university education, despite that essentially all of high school mathematics secretly uses it all the time! This makes mathematics look way crazier than it really is, which is a real shame.
So, let me try to explain it.
This will have the added benefit that you will know when to use $\equiv,$ in line with recommendation (2). Of course, you can always take recommendation (1) and just forget about $\equiv$ altogether... In any event, understanding this stuff is REALLY important.
Universal quantification.
Suppose we know that $\sin^2 \theta + \cos^2\theta = 1$.
Then obviously, we can deduce that $\sin^2(\theta+1)+\cos^2(\theta+1) = 1$.
(Think about why. This should be intuitively obvious. Don't read on until its obvious.)
Okay, suppose we instead know that $3x=2$. Can we deduce that $3(x+1) = 2$? Of course, we cannot. After all, the first statement is equivalent to $x=3/2$. The second statement is equivalent to $x=1/2$. Obviously, neither implies the other.
So what's really going on here?
Let me explain.
If someone says "we know that $3x=2$," then what they probably mean is exactly what they said. Basically, $x$ is understood to be some fixed but arbitrary real number, and we know that $3$ times that number equals $2$, or in other words that $3x=2$. We can then proceed to find $x$ (if we want).
But if someone says "we know that $\sin^2 \theta + \cos^2\theta = 1$," what they probably mean is: "we know that for each and every real number $\theta$, it holds that $\sin^2 \theta + \cos^2\theta = 1$." Obviously, we cannot proceed to find $\theta$ in this case, because we're talking about each and every possible $\theta$, not some specific $\theta$.
So if we want to be precise, if we want to denote things in a way that makes this difference in meanings clear, then obviously, we have to change our notation a bit.
The statement: "We know $3x=2$" remains unchanged.
But the other statement becomes: We know $$\mathop{\forall}_{\theta:\mathbb{R}}(\sin^2 \theta + \cos^2\theta = 1).$$
Let me explain how to read this. The symbol $\forall$ is verbalized "for all," or "for each," or "for every." A bit of terminology: we call $\forall$ the "universal quantifier." So the above pattern of symbols can be read: "for all $\theta$ in $\mathbb{R}$, it holds that $\sin^2 \theta + \cos^2\theta = 1$."
Often, we simply use words instead of symbols, so $$\mathop{\forall}_{x:X}[\mbox{blah blah}]$$ becomes: "for all $x$ of type $X$, blah blah."
Now please Google the words "function" and "predicate" if you're unfamiliar with these terms, because here comes the tricky part.
If you think about it, you'll see that if we're given a function $f : Y \leftarrow X$ and a predicate $P$ on $Y$, then from the statement $$\mathop{\forall}_{y:Y} P(y),$$ we can deduce $$\mathop{\forall}_{x:X} P(f(x)).$$ This explains how we got from $\sin^2 \theta + \cos^2\theta = 1$ to $\sin^2(\theta+1)+\cos^2(\theta+1) = 1.$ What's really going on is the following. We know that:
$$\mathop{\forall}_{\theta:\mathbb{R}}(\sin^2 \theta + \cos^2\theta = 1).$$
Now write $P(\theta)$ as shorthand for $\sin^2 \theta + \cos^2\theta = 1$. So we know that $$\mathop{\forall}_{\theta:\mathbb{R}}P(\theta)$$
Now define a function $f : \mathbb{R} \leftarrow \mathbb{R}$ as follows:
$$f(\theta) = \theta+1$$
Then we can deduce $$\mathop{\forall}_{\theta:\mathbb{R}}P(f(\theta)).$$
That is:
$$\mathop{\forall}_{\theta:\mathbb{R}}P(\theta+1).$$
That is:
$$\mathop{\forall}_{\theta:\mathbb{R}}(\sin^2 (\theta+1) + \cos^2(\theta+1) = 1).$$
which is what we were trying to show.
This explains why this pattern of reasoning works here, and not with $3x=2$. The notation $3x=2$ isn't shorthand for $$\mathop{\forall}_{x:\mathbb{R}}(3x=2),$$ since that would simply be false.
In summary, if we want our notation to make clear what the "rules of the game" are, then its a good idea to include the symbol $\forall$ or the phrase "for all" in our mathematical writing.
Returning to your original question.
Suppose you want to emphasize that the pattern of symbols $$\sin^2 \theta + \cos^2\theta = 1$$ is really a shorthand for the more correct statement $$\mathop{\forall}_{\theta:\mathbb{R}}(\sin^2 \theta + \cos^2\theta = 1).$$
Then, if you want, you can write $$\sin^2 \theta + \cos^2\theta \equiv 1$$ to emphasize this.
Example.
Let me finish by illustrating how and when to use $=$ versus $\equiv$, in line with recommendation (2).
Proposition. $1+\tan^2x \equiv \sec^2x$ over the real numbers.
Proof. Let $x$ denote a fixed but arbitrary real number. Then the following are equivalent.
- $1+\tan^2x = \sec^2x$
- $1+\sin^2 x / \cos^2 x = 1/\cos^2 x$
- $\cos^2x+\sin^2 x = 1$
- TRUE
Hence $1+\tan^2x \equiv \sec^2x$ over the real numbers. This completes the proof.
$$\tan\theta=1$$ is not an identity, it is an equation. Because it is true only for certain values of $\theta$.
$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ is an identity because whenever the members are defined, they are equal.
This is what is meant by "holds true for all the values of $\theta$".
If you prefer, there are no values of $\theta$ such that the two expressions differ.
Best Answer
Note that $\cos(x-y)=\cos x\cos y+\sin x\sin y$. This is obtained from the more familiar formula for $\cos(x+y)$ by replacing $y$ by $-y$.
Note also that $$a\cos\theta+b\sin\theta=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos\theta+ \frac{b}{\sqrt{a^2+b^2}}\sin\theta \right).$$
So if $\phi$ is the angle whose cosine is $\frac{a}{\sqrt{a^2+b^2}}$ and whose sine is $\frac{b}{\sqrt{a^2+b^2}}$, then in the formula above we can replace $\frac{a}{\sqrt{a^2+b^2}}$ by $\cos\phi$, and $\frac{b}{\sqrt{a^2+b^2}}$ with $\sin\phi$, and obtain $$a\cos\theta+b\sin\theta=\sqrt{a^2+b^2} \cos(\theta-\phi).$$
Remark: As to tricks and shortcuts, mostly it is a question of experience and practice. Already, I am sure, you recognize certain patterns and know how to exploit them. After a while, you will have used most of the common devices a dozen times, and then things get easy.