I should prove this trigonometric identity. I think I should get to this point : $\cos(3x) = 4\cos^3 x - 3\cos x $ But I don't have any idea how to do it (I tried solving $\cos(x+60^{\circ})\cos(x-60^{\circ})$ but I got nothing)
Asked
Active
Viewed 1.2k times
3
-
Please edit your question to show us the work you did so that we can see where you got stuck. – N. F. Taussig Feb 12 '15 at 18:47
-
@Triomath, You can try http://math.stackexchange.com/questions/1133590/evaluating-1-cos-frac-pi81-cos-frac3-pi8-1-cos-frac5-pi81/1133702#1133702 – lab bhattacharjee Feb 13 '15 at 05:41
3 Answers
5
Using sum of angles identity, \begin{align*} \cos x\cos (x-60^{\circ})&\cos(x+60^{\circ})\\ & = \cos x(\cos x\cos 60^{\circ}+\sin x\sin 60^{\circ})(\cos x\cos 60^{\circ}-\sin x\sin 60^{\circ})\\ \\ & = \cos x(\cos^2x\cos^260^{\circ}-\sin^2x\sin^260^{\circ})\\\\ & = \cos x\left(\frac{1}{4}\cos^2x-\frac{3}{4}\sin^2x\right)\\\\ & = \frac{1}{4}(\cos^3x-3\cos x(1-\cos^2x))\\\\ & = \frac{1}{4}(4\cos^3x-3\cos x) \end{align*}
Michael Rozenberg
- 194,933
Alijah Ahmed
- 11,609
-
-
You can use the identity $\cos(3x) = 4\cos^3x - \cos x$ that you stated in the problem. – N. F. Taussig Feb 12 '15 at 18:57
-
-
@Triomath You are correct. Alijah made a mistake when he evaluated the cosine and sine of $60^\circ$. He should have obtained $\cos x\left(\frac{1}{4}\cos^2x - \frac{3}{4}\sin^2x\right)$. – N. F. Taussig Feb 12 '15 at 19:03
-
@AlijahAhmed You made an error in evaluating the cosine and sine of $60^\circ$. Since $\cos(60^\circ) = 1/2$ and $\sin(60^\circ) = \sqrt{3}/2$, you should have obtained $\cos x\left(\frac{1}{4}\cos^2x - \frac{3}{4}\sin^2x\right)$. – N. F. Taussig Feb 12 '15 at 19:06
-
-
@N.F.Taussig. Yes indeed, I made a mistake. Many thanks for pointing it out - I have corrected my answer. – Alijah Ahmed Feb 12 '15 at 19:10
4
$ \cos (x-60^\circ ) \times \cos ( x + 60^\circ ) = \frac{1}{2} ( \cos (2x) + \cos 120^\circ) = \frac{1}{2} \cos (2x) - \frac{1}{4}$
$ (\frac{1}{2} \cos (2x) - \frac{1}{4}) \times \cos x = \frac{1}{4} ( \cos (3x) + \cos (x) ) - \frac{1}{4} \cos (x) = \frac{1}{4} \cos (3x) $
Tahir Imanov
- 459
-
@Triomath This solution makes use of the identity $$\cos\alpha\cos\beta = \frac{1}{2}[\cos(\alpha - \beta) + \cos(\alpha + \beta)]$$ – N. F. Taussig Feb 12 '15 at 19:08
0
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos (x-60^\circ )\cdot\cos ( x + 60^\circ )=\cos^2x-\sin^260^\circ=\cos^2x-\frac34$$
Now $\cos3x=4\cos^3x-3\cos x$
lab bhattacharjee
- 274,582