Use appropriate identities (sin, cos, tan)

Backlock · #1 $Old$ December 4th, 2008, 12:00 PM

Sup all, got untill 9PM GMT to get this done, and its the one i missed the lecture on so... Oh joy :P

Here goes..

Quote:

1) $cos3A=4cos^3A-3cosA$

2) $sin4A=4sinAcosA(cos^2A-sin^2A)$

Cheers (Y)

Edit: I think I have to work them out from these identitys, though Im not sure how.

$sin(A+B)=sinAcosB+sinBcosA$
$sin(A-B)=sinAcosB-sinBcosA$

$cos(A+B)=cosAcosB-sinAsinB$
$cos(A-B)=cosAcosB+sinAsinB$

$tan(A+B)=(tanA+tanB)/(1-tanAtanB)$ (?) - may be wrong with the +/-
$tan(A-B)=(tanA-tanB)/(1+tanAtanB)$ (?) - may be wrong with the +/-

nzmathman · #2 $Old$ December 4th, 2008, 12:25 PM

$\cos(3a)$

$= \cos(2a + a)$

$= \cos(2a) \, \cos(a) - \sin(2a) \, \sin(a)$
(expanding using cos(A+B) formula)

$= (2 \cos^2 (a) - 1) \, \cos(a) - 2 \sin(a) \, \cos(a) \, \sin(a)$
(using double angle formula for sin and cos)

$= 2cos^3(a) - \cos(a) - 2\sin^2(a) \, \cos(a)$

$= 2cos^3(a) - \cos(a) - 2\cos(a) \, [1 - \cos^2(a)]$

Now, I'm sure you can finish off...

nzmathman · #3 $Old$ December 4th, 2008, 12:30 PM

$\sin(4a)$

$= 2 \sin(2a) \cos(2a)$

$= 2 \times 2 \sin(a) \cos(a) \times [\cos^2(a) - \sin^2(a)]$

Which gives the answer required (only the double angle formula are needed for this question).

Soroban · #4 $Old$ December 4th, 2008, 12:32 PM

Hello, Backlock!

Of course, we need: . $\sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad\begin{Bmatrix}\sin^2\!\theta &=& 1 - \cos^2\!\theta \\ \cos^2\!\theta &=& 1 - \sin^2\!\theta \end{Bmatrix}$

We'll also need the Double-Angle Identities:
. . . . . $\sin2\theta \:=\:2\sin\theta\cos\theta \qquad\qquad\cos2\theta \:=\:\begin{Bmatrix}\cos^2\!\theta - \sin^2\!\theta \\ 2\cos^2\!\theta - 1 \\ 1 - 2\sin^2\!\theta \end{Bmatrix}$

Quote:

$\cos3A \:=\:4\cos^3\!A - 3\cos A$

$\cos3A \:=\:\cos(2A + A)$

. . . . . $= \;\underbrace{\cos2A}\cos A \quad-\quad \underbrace{\sin2A}\sin A$

. . $= \;\overbrace{(2\cos^2\!A - 1)}\cos A - \overbrace{2\sin A\cos A}\sin A$

. . . $= \;2\cos^3\!A - \cos A -\quad 2\underbrace{\sin^2\!A}\cos A$

. . . $= \;2\cos^3\!A - \cos A - 2\overbrace{(1 - \cos^2\!A)}\cos A$

. . . $= \;2\cos^3\!A - \cos A - 2\cos A + 2\cos^3\!A$

. . . $= \;4\cos^3\!A - 3\cos A$

Now try the other one yourself . . .

Backlock · #5 $Old$ December 4th, 2008, 12:41 PM

Cheers lads, think I got the nack of it now too