1

If $$\cos A+\cos B+\cos C=\sin A.+\sin B+\sin C=0$$ then prove that: $$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C)$$

My Attempt;

Here, $$\cos A+\cos B+\cos C=0$$ $$\cos A+\cos B=-\cos C$$ squaring on both sides, $$\cos^2A+2\cos A.\cos B+\cos^2B=\cos^2C$$

Again, $$\sin A+\sin B+\sin C=0$$ $$\sin A+\sin B=-\sin C$$ squaring on both sides $$\sin^2A+2\sin A.\sin B+\sin^2B=\sin^2C$$

Now,

$$L.H.S=\cos 3A+\cos 3B+\cos 3B$$ $$=4\cos^3A-3\cos A+4\cos^3B-3\cos B+4\cos^3C-3\cos C$$ $$=4(\cos^3A+\cos^3B+\cos^3C)-3(\cos A+\cos B+\cos C)$$ $$=4(\cos^3A+\cos^3B+\cos^3C)-3\times 0$$ $$=4(\cos^3A+\cos^3B+\cos^3C$$ Now, please help me to continue from.here.

NOTE: PLEASE DO NOT USE COMPLEX NUMBERS IN THE SOLUTION

pi-π
  • 8,352
  • Try to expand $(x+y+z)³$ and replace in the equation. – Rafael Aug 18 '16 at 07:53
  • Why do you ask not to use complex numbers (in shouting all-caps...) and accept the complex numbers based answer? Please be consistent in what you do — either the request is important, so the answer is invalid (even though concise and true), or the answer is acceptable and the requirement is void. – CiaPan Aug 31 '16 at 10:51
  • Actually, the requirement is more important. But no ome provided me the proof without complex numbers.. – pi-π Sep 01 '16 at 13:56

3 Answers3

3

HINT:

We have $$\sum(\cos A+i\sin A)=0$$

Using If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.,

$$\sum(\cos A+i\sin A)^3=3\prod(\cos A+i\sin A)$$

Can you take it from here?

See also: Clarification regarding a question

Community
  • 1
lab bhattacharjee
  • 279,016
3

Maybe you could try a complex approach (which is simpler ;-))

Let $z,u,v$ be three complex numbers: $z=\cos(A)+i\sin(A)$, $u=\cos(B)+i\sin(B)$, $v=\cos(C)+i\sin(C)$. then by hypothesis $z+u+w=0$. Hence $$0=(z+u+v)^3=z^3+u^3+v^3+3uv(u+v)+3uz(u+z)+3zv(z+v)+6zuv\\= z^3+u^3+v^3+3uv(-z)+3uz(-v)+3zv(-u)+6zuv=z^3+u^3+v^3-3zuv.$$ Therefore $z^3+u^3+v^3=3zuv$ which implies, after separating real and imaginary parts, $$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C)$$ and $$\sin 3A+\sin 3B+\sin 3C=3\sin(A+B+C).$$ So you get two (real) trigonometric relations at one stroke!

Robert Z
  • 147,345
1

Take a unit vector starting from the origin of a planar Cartesian coord. system at angle $A$.
Where is a geometric value of $\cos A$...?

Now add the next unit vector, starting at the end of the former one, at angle $B$ now.

Can you see what happens next? ...the sum of three cosines equal zero means the three-segment chain ends back at the $OY$ axis.
Similary, the sum of sines equal zero means the chain ends at the $OX$ axis.
Put together that means the chain returns to the origin...

And a closed 3-segment chain of equal-lengths segments is an equilateral triangle. So each angle differs from the previous one by $1/3$ of a full turn:

$$B = A + \frac 23 \pi,\ C = A + \frac 43 \pi$$ or $$B = A - \frac 23 \pi,\ C = A - \frac 43 \pi$$

Now you can substitute those B and C to your equality, expand functions, then reduce and see if you get an identity.

...or you can see that $$3B = 3A + 2\pi \text{ and } 3C = 3A + 4\pi$$ (or both pluses replaced with minuses) so $$\cos 3A + \cos 3B + \cos 3C = 3\cos 3A$$ and $$3\cos(A + B + C) = 3\cos(3A \pm 2\pi) = 3\cos 3A$$ which immediately implies the desired equality.

CiaPan
  • 13,884