Please help me with this trig question!

[thechicinnovation]

sin³a= (3sina-sin3a)/4

I don't understand the difference between the 3 before the sina and the 3 in between sin3a. Also, I don't what identities I should use to prove this problem.
a- alpha

 

[serds]

3sina means 3*sina and sin3a means sin(3a).

 

[thechicinnovation]

3sina means 3*sina and sin3a means sin(3a).

How should I open sin³​a?

 

[Bob Brown MSEE]

How should I open sin³​a?

sin³​a = (sin a)³ ​conventionally.

 

[thechicinnovation]

sin³​a = (sin a)³ ​conventionally.

I'm still confused, though. I don't know what I can do with the (sina)³ to make it into (3sin-sin3a)/ 4

 

[srmichael]

I'm still confused, though. I don't know what I can do with the (sina)³ to make it into (3sin-sin3a)/ 4

When proving trig identities, you are to make one side of the equation equal the other side of the equation. The question becomes "Which side do I choose to work on to make it look like the other side?" Teachers tell you to "work with the ugly side". Well, it is not always obvious which iside is "ugliest" and for that matter, it is not always true that working with the ugly side will lead you down the path of "solvation". In this problem, you wanted to work on $\displaystyle \sin^3x$ and quickly hit a road block. When that happens, see if it is easier to work on the other side. In this problem, it is better to work on the right side.

Hint: sin(3a) = sin(2a + a)

 

[thechicinnovation]

When proving trig identities, you are to make one side of the equation equal the other side of the equation. The question becomes "Which side do I choose to work on to make it look like the other side?" Teachers tell you to "work with the ugly side". Well, it is not always obvious which iside is "ugliest" and for that matter, it is not always true that working with the ugly side will lead you down the path of "solvation". In this problem, you wanted to work on $\displaystyle \sin^3x$ and quickly hit a road block. When that happens, see if it is easier to work on the other side. In this problem, it is better to work on the right side.

Hint: sin(3a) = sin(2a + a)

so.. 3sina-sin(2a+a)=3sina-sin²a-sina=2sina-sin2a=2sina-(2sina)(cosa)=2sina(1-cosa)/4.. what should I do from here?

 

[srmichael]

so.. 3sina-sin(2a+a)=3sina-sin²a-sina=2sina-sin2a=2sina-(2sina)(cosa)=2sina(1-cosa).. what should I do from here?

No. $\displaystyle sin(2a+a)\not=sin^2(a)+sin(a)$. Sin(2a) is the sine of an angle called 2a. Sin²a is the same as [sin(a)]² which is the square of the sine of angle "a".

$\displaystyle sin(2a+a)=sin(2a)cos(a)+cos(2a)sin(a)$=====> Does this formula look familiar?

$\displaystyle =2sin(a)cos(a)cos(a)+[1-2sin^2(a)]sin(a)$

$\displaystyle =2sin(a)cos^2(a)+[1-2sin^2(a)]sin(a)$

$\displaystyle =2sin(a)[1-sin^2(a)]+sin(a)-2sin^3(a)$

Can you take it from here?

 

[thechicinnovation]

!

No. $\displaystyle sin(2a+a)\not=sin^2(a)+sin(a)$. Sin(2a) is the sine of an angle called 2a. Sin²a is the same as [sin(a)]² which is the square of the sine of angle "a".

$\displaystyle sin(2a+a)=sin(2a)cos(a)+cos(2a)sin(a)$=====> Does this formula look familiar?

$\displaystyle =2sin(a)cos(a)cos(a)+[1-2sin^2(a)]sin(a)$

$\displaystyle =2sin(a)cos^2(a)+[1-2sin^2(a)]sin(a)$

$\displaystyle =2sin(a)[1-sin^2(a)]+sin(a)-2sin^3(a)$

Can you take it from here?

Yes, I got it! Thank you!