I have an exam tomorrow and it is highly likely that there will be a trig identity on it. To practice I tried this identity:

$$2 \sin 5x\cos 4x-\sin x = \sin9x$$

We solved the identity but we had to move terms from one side to another.

My question is: what are the things that you can and cannot do with trig identities?

And what things are must not when doing trig identities?

Thank you

  • $\begingroup$ This is not an identity (false for x=0. It is an equation whose roots are to be found. $\endgroup$ – marty cohen Jan 23 '13 at 18:33
  • $\begingroup$ Sorry about that. I accidentally added a "+" between $sin5x$ and $cos4x$ $\endgroup$ – Jeel Shah Jan 23 '13 at 18:42

Those are very general questions: what "can you do" and what "can't you do"...in terms of using trig identities:

What you can do is use an identity to replace one expression with its strict equivalent, as determined by the identity in question. It's simply grounded in "substituting one expression for its equivalent", which is legitimate beyond just its handiness with utilizing trig identities to transform one expression to an equivalent expression.

You can algebraically manipulate equations involving trigonometric expressions in any way that is "legal" for manipulating any equation.

So for a simple example, ${1-\sin^2\theta} = \cos^2 \theta$ because we know the identity:

$$\sin^2\theta + \cos^2\theta = 1.$$ We do not change this equivalence when we manipulate it to get: $$\sin^2\theta + \cos^2\theta = 1 \iff \cos^2\theta = 1 - \sin^2\theta$$

In the identity you solved, you can use:

$$\sin c+\sin d=2\sin\frac{c+d}2\cos\frac{c-d}2$$

Applied here: $$\sin9x+\sin x=2\sin\frac{9x+x}2\cos\frac{9x-x}2=2\sin5x\cos4x$$

$$2\sin a\cos b=\sin(a+b)+\sin(a-b)$$

Applied here: $$2\sin5x\cos4x=\sin(5x+4x)+\sin(5x-4x)=\sin9x+\sin x$$

  • $\begingroup$ I can move terms from one side to another as long as both side are equal right? $\endgroup$ – Jeel Shah Jan 23 '13 at 18:16
  • $\begingroup$ Yes, indeed! Just as I took the identity above, subtracted $\sin^2 \theta$ from both sides of the identity $\sin^2\theta + \cos^2\theta = 1$ $\endgroup$ – Namaste Jan 23 '13 at 18:23
  • $\begingroup$ Are you asking specifically about the identity in question (that you solved)? Or about trig identities in general? I took your question to refer to "in general". If you don't find my answer helpful, I can delete it. $\endgroup$ – Namaste Jan 23 '13 at 19:02
  • $\begingroup$ Yes, my question was general and yes your answer is helpful. $\endgroup$ – Jeel Shah Jan 23 '13 at 19:03

Hint: We have $\sin 5x\cos 4x+\cos 5x\sin 4x=\sin 9x$. Replace one of the $\sin 5x\cos 4x$ by $\sin 9x-\cos 5x\sin 4x$. Leave the other one alone. Something nice will happen.

You will end up having used the basic addition and subtraction laws $$\sin(a\pm b)=\sin a\cos b\pm \cos a\sin b.$$

  • $\begingroup$ In fact, the "subtraction" laws are just a matter of substituting $-b$ into the "addition" laws, which is why I don't consider them worth mentioning. $\endgroup$ – pre-kidney Jan 23 '13 at 19:14
  • $\begingroup$ @pre-kidney: True, but in this context it serves to extend the hint. $\endgroup$ – André Nicolas Jan 23 '13 at 20:08

The intended techniques all follow from using the sine and cosine addition formulas and normalization, which you should have seen before.

However, I wanted to point out that a more unified approach to the general problem of proving trig. identities is to work in the complex plane. For instance $e^{ix}=\cos(x)+i\sin(x)$ allows you to derive identities such as $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ and $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$. Let $w=e^{ix}$. Then you can turn your identity into an equivalent "factorization problem" for a sparse polynomial in $w$, of degree $18$.

Try it out!

  • 1
    $\begingroup$ Working in the complex plane is not an option because we haven't learned that yet. $\endgroup$ – Jeel Shah Jan 23 '13 at 18:15
  • $\begingroup$ Then as I said, all your techniques will follow from addition formulas and normalization. Also, I checked your "identity", and it does not hold. Is there a typo somewhere? In case you meant "equation", not "identity", I checked for roots and saw that it didn't have any computable ones, either. $\endgroup$ – pre-kidney Jan 23 '13 at 18:26

$$\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$$

So, $$\sin9x+\sin x=2\sin\frac{9x+x}2\cos\frac{9x-x}2=2\sin5x\cos4x$$

$$2\sin A\cos B=\sin(A+B)+\sin(A-B)$$

So, $$2\sin5x\cos4x=\sin(5x+4x)+\sin(5x-4x)=\sin9x+\sin x$$

  • $\begingroup$ Observe that the two formulae are related and one can be easily derived from the other. $\endgroup$ – lab bhattacharjee Jan 23 '13 at 19:09

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