Trig Identities: 10(4Sin2A+SinB)+2(4Sin2A+SinB)2+0.5(4Sin2A+SinB)3

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BWJohnson

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I have a problem which I have to expand and need to remove any powers or product terms similar to this, I know that trig idents will be used.

10(4SinA+SinB)+2(4SinA+SinB)2+0.5(4SinA+SinB)3

I understand that the first step could be to expand all of the brackets:

10(4SinA+SinB)+2(4SinA+SinB)(4SinA+SinB)+0.5(4SinA+SinB)(4SinA+SinB)(4SinA+SinB)
10(4SinA+SinB)+2(16Sin2A+8SinASinB+Sin2B)+0.5(16Sin2A+8SinASinB+Sin2​B)(4SinA+SinB)
10(4SinA+SinB)+2(16Sin2A+8SinASinB+Sin2B)+0.5(64Sin3A+48SinA2SinB+12SinASin2B+Sin3B)

Then I will multiply all of the terms in the bracket by the outside figure

40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these expressions do not need any trig idents
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these expressions need the Sin2A=1/2(1-cos2A)
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these expressions need the Sin3A=3/4SinA+1/4Sin3A
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -this expression needs the Sin(A)Sin(B)=1/2(cos(A-B)-cos(A+B)
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these are the 2 im struggling with a bit!!!!

I can see that there is a square and a Product, do i apply the suare rule and then the COSSIN Product Rule for example;

6SinASin2B
6((SinA)(1/2(1-cos2B)) Apply the rule
6((SinA)(1/2-1/2cos2B)) Multiply terms within the answer by 1/2
6(1/2SinA-1/2SinACosB) Expand the brackets
3SinA-3SinACosB Multiply in 6 and this term now has no trig ident
3SinA-3(1/2(cos(A-B)-cos(A+B)) Apply the SinACosB rule
3SinA-3/2(cos(A-B)-cos(A+B)) Multiply in 3
3SinA-3/2cosA+3/2cosB-3/2cosA-3/2cosB All terms are now complete however they also all cancel out when summed together giving 0. This is not right as far as im aware, I have used Mathcad which gives a different final answer.

What is the correct process for applying the laws to this type of expression.
 
Last edited:
BWJohnson said:
I have a problem which I have to expand and need to remove any powers or product terms similar to this, I know that trig idents will be used.

10(4Sin2A+SinB)+2(4Sin2A+SinB)2+0.5(4Sin2A+SinB)3

I understand that the first step could be to expand all of the brackets:

10(4SinA+SinB)+2(4SinA+SinB)(4SinA+SinB)+0.5(4SinA+SinB)(4SinA+SinB)(4SinA+SinB)You have lost the ^2 in 4sinA terms within the brackets
10(4SinA+SinB)+2(16Sin2A+8SinASinB+Sin2B)+0.5(16Sin2A+8SinASinB+Sin2​B)(4SinA+SinB)
10(4SinA+SinB)+2(16Sin2A+8SinASinB+Sin2B)+0.5(64Sin3A+48SinA2SinB+12SinASin2B+Sin3B)

Then I will multiply all of the terms in the bracket by the outside figure

40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these expressions do not need any trig idents
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these expressions need the Sin2A=1/2(1-cos2A)
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these expressions need the Sin3A=3/4SinA+1/4Sin3A
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -this expression needs the Sin(A)Sin(B)=1/2(cos(A-B)-cos(A+B)
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these are the 2 im struggling with a bit!!!!

I can see that there is a square and a Product, do i apply the suare rule and then the COSSIN Product Rule for example;

6SinASin2B
6((SinA)(1/2(1-cos2B)) Apply the rule
6((SinA)(1/2-1/2cos2B)) Multiply terms within the answer by 1/2
6(1/2SinA-1/2SinACosB) Expand the brackets
3SinA-3SinACosB Multiply in 6 and this term now has no trig ident
3SinA-3(1/2(cos(A-B)-cos(A+B)) Apply the SinACosB rule
3SinA-3/2(cos(A-B)-cos(A+B)) Multiply in 3
3SinA-3/2cosA+3/2cosB-3/2cosA-3/2cosB All terms are now complete however they also all cancel out when summed together giving 0. This is not right as far as im aware, I have used Mathcad which gives a different final answer.

What is the correct process for applying the laws to this type of expression.
Click to expand...
See red comment
 
My mistake.

my mistake. I removed it from the working out but neglected to remove it from the original expression.

so to be clear the original expression should be

10(4SinA+SinB)+2(4SinA+SinB)2+0.5(4SinA+SinB)3

I did all the working out for this example quickly so may have made some simple errors however the last question still stands. It appears to be a process error.
 
BWJohnson said:
I have a problem which I have to expand and need to remove any powers or product terms similar to this, I know that trig idents will be used.

10(4SinA+SinB)+2(4SinA+SinB)2+0.5(4SinA+SinB)3

I understand that the first step could be to expand all of the brackets:

10(4SinA+SinB)+2(4SinA+SinB)(4SinA+SinB)+0.5(4SinA+SinB)(4SinA+SinB)(4SinA+SinB)
10(4SinA+SinB)+2(16Sin2A+8SinASinB+Sin2B)+0.5(16Sin2A+8SinASinB+Sin2 B)(4SinA+SinB)
10(4SinA+SinB)+2(16Sin2A+8SinASinB+Sin2B)+0.5(64Sin3A+48SinA2SinB+12SinASin2B+Sin3B)

Then I will multiply all of the terms in the bracket by the outside figure

40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these expressions do not need any trig idents
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these expressions need the Sin2A=1/2(1-cos2A)
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these expressions need the Sin3A=3/4SinA+1/4Sin3A
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -this expression needs the Sin(A)Sin(B)=1/2(cos(A-B)-cos(A+B)
40SinA+10SinB+32Sin2A+16SinASinB+2Sin2B+32Sin3A+24SinA2SinB+6SinASin2B+0.5Sin3B -these are the 2 im struggling with a bit!!!!

I can see that there is a square and a Product, do i apply the suare rule and then the COSSIN Product Rule for example;

6SinASin2B
6((SinA)(1/2(1-cos2B)) Apply the rule
6((SinA)(1/2-1/2cos2B)) Multiply terms within the answer by 1/2
6(1/2SinA-1/2SinACosB) Expand the brackets
3SinA-3SinACosB Multiply in 6 and this term now has no trig ident
3SinA-3(1/2(cos(A-B)-cos(A+B)) Apply the SinACosB rule
3SinA-3/2(cos(A-B)-cos(A+B)) Multiply in 3
3SinA-3/2cosA+3/2cosB-3/2cosA-3/2cosB This is where the error occurred - You can not multiply the terms within the bracket by that outside
All terms are now complete however they also all cancel out when summed together giving 0. This is not right as far as im aware, I have used Mathcad which gives a different final answer.

What is the correct process for applying the laws to this type of expression.
Click to expand...

See Comments in RED for the error. The 3SinA-3/2cos(A-B)+3/2cos(A+B)) is the final product of the expression
 
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