Question

how do you solve 4sin5θcos5θ in terms of single trigonometric function?

Answer 1

Use sin2x = 2sinxcosx
Put x= 5θ in this case
4sin5θcos5θ
= 2 ( 2sin5θcos5θ)
= ...?

Answer 2

why did you put 2(sin5θcos5θ)?

Answer 3

For you to observe easier and apply that identity.

Answer 4

so it becomes 2(sin5θcos5θ)= 2sin10θcos10θ?

Answer 5

Nope..
2sinxcosx = sin2x

Answer 6

how did cosx disappear?

Answer 7

It's an identity....
sin 2x
= sin (x+x)
= sinx cosx + cosx sinx
= 2 sinx cosx

Answer 8

i get the identity part. but is it possible to show me in steps after 2(sin5θcos5θ)?

Answer 9

As I've told you, put x=5θ into the identity
sin2x = 2sinxcosx
For the right, you get something familiar, which is a part of your question
For the left, it's something you want. What do you get for the left?

Answer 10

sin10θ ?

Answer 11

Yes.
But for the question, you have
4 sin5θ cos5θ
So, you'll have
2 (2sin5θ cos 5θ)
^ Can you simplify it?

Answer 12

I dont know how to simplify this....

Answer 13

1. simplify this: (2sin5θ cos 5θ)
2. Multiply the result you get by 2

What do you get for the first step?

Answer 14

1) sin2a

Answer 15

i mean sin25θ

Answer 16

Why is it 25θ??

Answer 17

so 4sin5θ cos5θ = sin2x and x=5θ so sin2(5θ)=sin10θ ...isn't it done?

Answer 18

50% correct.
4sinx cosx = 2sin2x
Put x = 5θ into the above identity.

Answer 19

Oh because 2sinx cos x = sin2x and here, its 4sinx cos x so we put 2sin2x??

Answer 20

Yes.

Answer 21

Ah thank you very much!! makes much more sense

Answer 22

Is there such a question that it can be like (any number)sinxcosx?

Answer 23

I'm not sure...
but 2sinxcosx = sin2x
If you have sinxcosx, you can rewrite it as 1/2 (sin2x)
Btw, remember to put x = 5θ back to your question!

Answer 24

so if it was sinxcosx= 1/2sinx?

Answer 25

No!
sinxcosx= 1/2sin2x

Answer 26

It's double angle formula!

Answer 27

oh Yeah i made a mistake, i meant by 1/2sin2x

Answer 28

Yes. \(sinxcosx = \frac{1}{2}sin2x\)

Answer 29

thank you!!

Answer 30

Welcome!