# Another trig question

1. Apr 11, 2014

### Attis

1. The problem statement, all variables and given/known data
show that sin4x + 2sin2x = 8sinxcos^3x

2. Relevant equations
sin2x =2sinxcosx

3. 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!
??

2. Apr 11, 2014

### Tanya Sharma

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

sin4x = 2sin2xcos2x

3. Apr 11, 2014

### Attis

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?)

4. Apr 11, 2014

### Tanya Sharma

Yes

Last edited: Apr 11, 2014
5. Apr 11, 2014

### Attis

2sinAcosA?

Or is this some sort of a trick question?

6. Apr 11, 2014

### Tanya Sharma

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 ?

7. Apr 11, 2014

### Attis

Ah!! I see. Thanks!
sin2*2x= 2sin2xcos2x.
I´ll carry on now and see if I get anywhere.

8. Apr 11, 2014

### Attis

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...

9. Apr 11, 2014

### Tanya Sharma

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

10. Apr 11, 2014

### Attis

2sin2x(cos2x + 1)?

11. Apr 11, 2014

### Tanya Sharma

Good...

1+cos2x = ?

12. Apr 11, 2014

### Attis

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

13. Apr 11, 2014

### Tanya Sharma

Correct...Is there some other way of expressing this expression ? Can a couple of terms be combined ?

14. Apr 11, 2014

### Attis

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)

?

15. Apr 11, 2014

### Tanya Sharma

Right...

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

Last edited: Apr 11, 2014
16. Apr 11, 2014

### Attis

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?

17. Apr 11, 2014

### Tanya Sharma

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 .

18. Apr 11, 2014

### Attis

Ok. I´ll do my best. Thanks once again!