Proof expression: $$\cos(80^\circ)\sin(70^\circ)\cos(60^\circ)\sin(50^\circ)=\frac{1}{16}$$
I tried in different ways, but I always get $\sin(20^\circ)$ in expression, which I can't simplify. What is the fastest way to prove this?
Asked
Active
Viewed 847 times
-2
-
1Use https://math.stackexchange.com/questions/1145406/prove-that-cosx-cosx-60-cosx60-1-4cos3x – lab bhattacharjee Dec 16 '17 at 14:45
2 Answers
4
$$\cos80°\sin70°\cos60°\sin50°=\frac{8\sin20^{\circ}\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}}{16\sin20^{\circ}}=$$ $$=\frac{\sin160^{\circ}}{16\sin20^{\circ}}=\frac{1}{16}$$
Michael Rozenberg
- 194,933
-
-
-
How did you simplify in the first step? That you got: $\frac{8\sin(20)\cos(20)\cos(40)\cos(80)}{16\sin20}$ – VLC Dec 16 '17 at 15:09
-
$\sin70^{\circ}=\cos20^{\circ}$, $\sin50^{\circ}=\cos40^{\circ}$ and $\cos60^{\circ}=\frac{1}{2}$. – Michael Rozenberg Dec 16 '17 at 15:15
-
-
-
-
$8\sin20^{\circ}\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}=4\sin40^{\circ}\cos40^{\circ}\cos80^{\circ}=2\sin80^{\circ}\cos80^{\circ}=\sin160^{\circ}.$ – Michael Rozenberg Dec 16 '17 at 15:35
-
Thanks a lot, where could I learn to proof expressions this way, like you did. I tried to proof with addition formulas and I've written 2 A4 pages of simplification and still didn't proof anything. – VLC Dec 16 '17 at 15:48
-
You are welcome! Just solve math problems always and It will come. – Michael Rozenberg Dec 16 '17 at 15:55
1
As @labbhattacharjee suggested, use the link with $x=20^{\circ}$ so that $$\cos(20^{\circ})\cos(-40^{\circ})\cos(80^{\circ})=\sin(70^{\circ})\sin(50^{\circ})\cos(80^{\circ})=\frac14 \cos(3\cdot 20^{\circ})=\frac18$$ Hence $$\cos(80^\circ)\sin(70^\circ)\cos(60^\circ)\sin(50^\circ)=\frac12 \cdot \frac18 =\frac{1}{16}$$ as required.
ə̷̶̸͇̘̜́̍͗̂̄︣͟
- 27,005