1
$\begingroup$

How do I simplify the expression of $\cos(2x+3x)\cos x+\sin(3x)\cos\left(\frac x2+x\right)$ ?

$\endgroup$
  • $\begingroup$ I've fixed tex, please check that I didn't modify anything. $\endgroup$ – Davide Giraudo May 26 '12 at 19:54
  • 3
    $\begingroup$ Why did you write it as $2x+3x$ instead of $5x$? $\endgroup$ – Phira May 26 '12 at 20:05
  • 2
    $\begingroup$ @JoeL. is it possible to use (cos(2x)+cos(3x))⋅cosx ?? i calculated and found (for example,in degrees) cos(15)+cos(30) != cos(45) :( $\endgroup$ – DrStrangeLove May 26 '12 at 20:06
  • $\begingroup$ A simple simplification is $\,\cos (2x+3x)=\cos 5x\,$...is there a typo here? $\endgroup$ – DonAntonio May 26 '12 at 20:11
  • $\begingroup$ hat do you want to simplify it into? It can be done cleanly into a function of $\cos(x/2)$ and $\sin(x/2)$. $\endgroup$ – ncmathsadist May 26 '12 at 20:12
2
$\begingroup$

$$ \begin{align} \cos\alpha\cos\beta - \sin\alpha\sin\beta & = \cos(\alpha+\beta) \\ \cos\alpha\cos\beta + \sin\alpha\sin\beta & = \cos(\alpha-\beta) \end{align} $$ Adding left sides and adding right sides gives $$ 2\cos\alpha\cos\beta = \cos(\alpha+\beta)+\cos(\alpha-\beta) $$ so $$ \cos\alpha\cos\beta = \frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}. $$ A similar thing handles a product of a sine and a cosine.

Later edit: "Dr. Strangelove" doesn't seem to be getting excited about this answer, so I'll add a bit more. The proposed identity includes $\cos(5x)\cos x$. Let $5x$ be $\alpha$ and $x$ be $\beta$ in the identity above. We get $$ \cos(2x+3x)\cos x=\cos(5x)\cos x = \frac{\cos(5x+x) + \cos(5x-x)}{2} = \frac{\cos(6x)+\cos(4x)}{2}. $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.