# Another trig question

## Homework Statement

show that sin4x + 2sin2x = 8sinxcos^3x

sin2x =2sinxcosx

## The Attempt at a Solution

I started out by letting sin4x = sin(2x*2) so that I could plugg in sin2x = 2sinxcosx in the equation.
sin(2x*2) +2sin2x =
sin2*sin2x +2sin2x =
sin2 * (2sinxcosx) + 2*sin2x =
sin2* (2sinxcosx) + 2* (2sinxcosx)
and I end up with 8sinxcosx
which is wrong!
??

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sin(2x*2) +2sin2x =
sin2*sin2x +2sin2x =
sin2 * (2sinxcosx) + 2*sin2x =
sin2* (2sinxcosx) + 2* (2sinxcosx)
and I end up with 8sinxcosx
which is wrong!
??
sin(ab) ≠ (sina)(sinb) i.e sin4x≠(sin2)(sin2x)

sin4x = 2sin2xcos2x

sin(ab) ≠ (sina)(sinb) i.e sin4x≠(sin2)(sin2x)

sin4x = 2sin2xcos2x
Ok, did you derive that from sin2x = 2sinxcosx?
In that case, I don´t understand why sin4x = 2sin2xcos2x and not 4sin2xcos2x (i.e twice as much as sin2x = 2sinxcosx?)

Ok, did you derive that from sin2x = 2sinxcosx?
Yes

In that case, I don´t understand why sin4x = 2sin2xcos2x and not 4sin2xcos2x (i.e twice as much as sin2x = 2sinxcosx?)

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Yes

2sinAcosA?

Or is this some sort of a trick question?

Or is this some sort of a trick question?
Not at all . Just trying to make you understand the formula 2sinAcosA?
sin2A = 2sinAcosA . Now replace A with 2x on both the sides .What do you get ?

Not at all . Just trying to make you understand the formula sin2A = 2sinAcosA . Now replace A with 2x on both the sides .What do you get ?
Ah!! I see. Thanks!
sin2*2x= 2sin2xcos2x.
I´ll carry on now and see if I get anywhere.

Somehow it keeps on getting more and more complicated. Do you have any idea on how I could keep it "simple"?

sin4x + 2sin2x = 8sinxcos^3x
on the left hand side:
2sin2xcos2x + 2sin2x =
2*2sinxcosx(cos^2x - sin^2x) + 2*2sinxcosx =
4sinxcos^3x - 2sin^3xcosx + 4sinxcosx

It really doesn´t feel right...

2sin2xcos2x + 2sin2x
Is there something common in the two terms ? If yes ,take out the common factor .What do you get ?

Is there something common in the two terms ? If yes ,take out the common factor .What do you get ?
2sin2x(cos2x + 1)?

Good...

1+cos2x = ?

1 + cos^2x - sin^2x
?

1 + cos^2x - sin^2x
?
Correct...Is there some other way of expressing this expression ? Can a couple of terms be combined ?

There are two options:
1) 1 + cos^2x -sin^2x =
1+cos^2x - (1-cos^2x)=2cos^2x

2) 1+(1-sin^2x)-sin^2x=
1+1 - sin^2x -sin^2x = 2-2sin^2x = 2(1-sin^2x)

?

There are two options:
1) 1 + cos^2x -sin^2x =
1+cos^2x - (1-cos^2x)=2cos^2x
Right...

Now put sin2x = 2sinxcosx and (1+cos2x) = 2cos2x in the expression you have in post#10.

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• 1 person
Right...

Now put sin2x = 2sinxcosx and (1+cos2x) = 2cos2x in the expression you have in post#10.
Perfect! Now I got it.
Thanks a lot!! you´ve been a massive help.
Do you have any general tips on how to solve such problems?

Remembering trigonometric formulas and practicing variety of problems of increasing complexity is the key .If you remember the formulas then it will help you in manipulating the trigonometric expressions .

You should be comfortable with double ,triple angle formulas .How to convert from one form to other .

Cos2x = cos2x-sin2x = 1-2sin2x = 2cos2x-1 .

sinx = √[(1-cos2x)/2] ; cosx = √[(1+cos2x)/2]

All the above are various ways of writing the same thing .With practice you will recognize which form to use .

Remembering trigonometric formulas and practicing variety of problems of increasing complexity is the key .If you remember the formulas then it will help you in manipulating the trigonometric expressions .

You should be comfortable with double ,triple angle formulas .How to convert from one form to other .

Cos2x = cos2x-sin2x = 1-2sin2x = 2cos2x-1 .

sinx = √[(1-cos2x)/2] ; cosx = √[(1+cos2x)/2]

All the above are various ways of writing the same thing .With practice you will recognize which form to use .
Ok. I´ll do my best. Thanks once again!