Using the identity $cos (A + B) = cos A cos B - sin A sin B$ $cos(A+B) = cosAcosB - sinAsinB$ , how do you prove that $\frac{1}{4} cos (3 A) = {cos}^{3} A - \frac{3}{4} cos A$ $1/4cos(3A) = cos^3A - 3/4cosA$ ?

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Using the identity $cos (A + B) = cos A cos B - sin A sin B$ $cos(A+B) = cosAcosB - sinAsinB$ , how do you prove that $\frac{1}{4} cos (3 A) = {cos}^{3} A - \frac{3}{4} cos A$ $1/4cos(3A) = cos^3A - 3/4cosA$ ?

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Answer 1 · 2016-06-12T23:53:12

I'm going to work from the right hand side. For $cos 3 A$ $cos3A$ , we know that:

$cos 3 A = cos (A + 2 A) = cos A cos 2 A - sin A sin 2 A .$ $color(green)(cos3A = cos(A + 2A) = cosAcolor(highlight)(cos2A) - sinAcolor(highlight)(sin2A)).$

Next, recall that:

$sin 2 A = sin A cos A + cos A sin A = 2 sin A cos A$ $color(highlight)(sin2A) = sinAcosA + cosAsinA = color(highlight)(2sinAcosA)$
$cos 2 A = cos A cos A - sin A sin A = {cos}^{2} A - {sin}^{2} A$ $color(highlight)(cos2A) = cosAcosA - sinAsinA = color(highlight)(cos^2A - sin^2A)$

So, we now have:

$cos 3 A = cos A ({cos}^{2} A - {sin}^{2} A) - sin A (2 sin A cos A)$ $cos3A = cosA(color(highlight)(cos^2A - sin^2A)) - sinA(color(highlight)(2sinAcosA))$

$= {cos}^{3} A - {sin}^{2} A cos A - 2 {sin}^{2} A cos A$ $= cos^3A - sin^2AcosA - 2sin^2AcosA$

$= {cos}^{3} A - 3 {sin}^{2} A cos A$ $= cos^3A - 3color(red)(sin^2)AcosA$

What I'm shooting for at this point is the result you get on the left hand side when you multiply the right hand side by $4$ $4$ : $4 {cos}^{3} A - 3 cos A$ $color(purple)(4cos^3A - 3cosA)$ . That requires all functions to be $cos$ $cos$ here.

So, we have to reference the identity that you definitely know: ${sin}^{2} A + {cos}^{2} A = 1$ $color(red)(sin^2A + cos^2A = 1)$ . This gives us:

$= {cos}^{3} A - 3 (1 - {cos}^{2} A) cos A$ $= cos^3A - 3(color(red)(1 - cos^2A))cosA$

$= {cos}^{3} A - 3 (cos A - {cos}^{3} A)$ $= cos^3A - 3(cosA - cos^3A)$

$= {cos}^{3} A - 3 cos A + 3 {cos}^{3} A$ $= cos^3A - 3cosA + 3cos^3A$

$= 4 {cos}^{3} A - 3 cos A$ $= color(purple)(4cos^3A - 3cosA)$

Almost there! Finally, since this was equal to $cos 3 A$ $cos3A$ , we have to divide by $4$ $4$ to get:

$\frac{1}{4} cos 3 A = \frac{1}{4} \cdot [4 {cos}^{3} A - 3 cos A]$ $1/4cos3A = 1/4*[4cos^3A - 3cosA]$

$\Rightarrow \frac{1}{4} cos 3 A = {cos}^{3} A - \frac{3}{4} cos A$ $=> color(blue)(1/4cos3A = cos^3A - 3/4cosA)$

Answer 2 · 2016-06-13T18:52:48

Using only the given identity of $cos (A + B)$ $cos(A+B)$ the other given identity can be proved as follows.

Explanation:

General Discussion

The identity to be assumed is

$cos(A+B)=cosAcosB-sinAsinB...(1)$

As this is valid for all real values of A and B , we can put convenient values to get different identities as follows.

Method - I( A Tricky one)

Putting $B= -B$ in equation (1)

$cos(A-B)=cosAcos(-B)-sinAsin(-B)$

$=>cos(A-B)=cosAcosB+sinAsinB…(1.1)$

Adding (1) and (1.1) we get

$cos(A+B)+cos(A-B)=2cosAcosB…(1.2)$

Now Putting $B=2A$ in (1.2)

$cos(A+2A)+cos(A-2A)=2cosAcos2A$

$cos3A+cosA=2cosAcos2A$

$cos3A=cosA(2cos2A-1)…(1.3)$

Putting $B=A$ in (1.2)

$cos(A+A)+cos(A-A)=2cosAcosA$

$=>cos2A+cos0=2cos^2A$

$cos2A=2cos^2A-1…(1.4)$

Finally combining (1.4) with (1.3) we get

$cos3A=cosA(2(2cos^2A-1)-1)$

$=>cos3A=4cos^3A-3cosA$

Re-arrenging and deviding bothsides by 4
$cos^3A-3/4cosA=1/4cos3A$
Proved

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Method-II (A long one)

Putting $B=A$ in equation(1)

$cos(A+A)=cosAcosA-sinAsinA$

$cos2A=cos^2A-sin^2A....(2)$

Putting $B=90+A$ in equation(1)

$cos(A+90+A)=cosAcos(90+A)-sinAsin(90+A)$

$-sin2A=-sinAcosA-sinAcosA$

$=>sin2A=2sinAcosA...(3)$

Putting $B=-A$ in equation(1)

$cos(A-A)=cosAcos(-A)-sinAsin(-A)$

$=>cos^2A+sin^2A=1...(4)$

Now finally putting $B=2A$ in equation (1)

$cos(A+2A)=cosAcos2A-sinAsin2A$

inserting the relation (3)

$cos3A=cosAcos2A-2sinAsinAcosA$

$=cosA(cos2A-2sin^2A)$

Inserting relation (2)

$cos3A=cosA(cos^2A-sin^2A-2sin^2A)$

$=cosA(cos^2A-3sin^2A)$

$=cosA(4cos^2A-3cos^2A-3sin^2A)$

$=cosA(4cos^2A-3(cos^2A+sin^2A)$

$=cosA(4cos^2A-3*1)" " relation(4)$

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

Re-arrenging and deviding bothsides by 4

$:.cos^3A-3/4cosA=1/4cos3A$

Proved

Answer 3 · 2016-06-14T12:41:39

RHS is
$1/4cos(3A) = 1/4cos(2A+A)$
Using the given expression
$=> 1/4(cos (2A)cos(A) - sin(2A)sinA)$

Substituting the known expression for $sin (2A)=2sinA cos A$
Using given expression again for
$cos(2A)=cos (A+A)=cos^2A-sin^2A$
and writing all terms in $"cosine"$ using $sin^2A+cos^2A=1$ we obtain
$1/4((2cos^2A-1)cosA - 2sin^2AcosA)$

$=> 1/4(2cos^3A - cosA - 2(1-cos^2A)cosA)$
$= 1/4(2cos^3A - cosA - 2cosA + 2cos^3A)$
$=>1/4( 4cos^3A - 3cosA)$
$=> cos^3A - 3/4 cosA=$ LHS

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Using the identity $cos (A + B) = cos A cos B - sin A sin B$ $cos(A+B) = cosAcosB - sinAsinB$ , how do you prove that $\frac{1}{4} cos (3 A) = {cos}^{3} A - \frac{3}{4} cos A$ $1/4cos(3A) = cos^3A - 3/4cosA$ ?

3 Answers

Explanation:

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Explanation:

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Using the identity $cos (A + B) = cos A cos B - sin A sin B$ $cos(A+B) = cosAcosB - sinAsinB$ , how do you prove that $\frac{1}{4} cos (3 A) = {cos}^{3} A - \frac{3}{4} cos A$ $1/4cos(3A) = cos^3A - 3/4cosA$ ?