Question

solve sinθ+sin3θ=sin2θ for 0<θ<180

Answer 1

Hello @jimswig88 . Let me first make sure that do you know the formula for \(\sin 3 \theta \) and \(\sin 2 \theta\)

Answer 2

Is sin3θ =sin2θcosθ+cos2θsin? or is that a mistake

Answer 3

wait! I think I have to re-login here

Answer 4

I am back .

Answer 5

\[\sin 3\theta =\sin (2\theta+\theta)=\sin 2\theta \cos \theta+\cos 2\theta \sin \theta\]

Answer 6

Write : \(\large{\sin 3 \theta = \sin (2\theta + \theta )}\)

Answer 7

And then use : \(\large{\color{blue}{\sin(A+B)}}\) . I hope you know the formula for sin(A+B) ?

Answer 8

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Answer 9

@jimswig88 can u tell me the formula for : \(\large{\color{red}{\sin(A+B)}}\)

Answer 10

sinAcosB+sinBcosA?

Answer 11

Right . So what will be : \(\large{\color{grey}{\sin(2\theta + \theta)}}\) ?

Answer 12

Unsure were to go from hre

Answer 13

See : \(\large{\color{blue}{ \sin(2\theta +\theta)} = \color{red}{\sin 2\theta \cos \theta + \cos 2\theta \sin \theta }}\)
right?

Answer 14

Oh ok !makes sense

Answer 15

Now , we have : \(\large{\sin \theta + \sin 3\theta = \sin 2\theta}\)

What if i put the formula up there for sin 2 theta also

Answer 16

Double angle formula yes?

Answer 17

yes. Let us both try to do that .

Answer 18

2sinθcosθ?

Answer 19

yep

Answer 20

oh good stuff. So i have sinθ+sin2θcosθ+cos2θsinθ=2sinθcosθ. How is that looking?

Answer 21

Oh that looks scary :P

Answer 22

Ha yes it does

Answer 23

I got it :
we have \(\large{\sin \theta + \sin 3 \theta = \sin 2\theta}\)

why dont' use sin A + sin B formula ?

Answer 24

Do you know the formula for sin A + sin B ?

Answer 25

I can not find it in my notes. I know its simple arrrghhh!

Answer 26

\[\large{\color{blue}{\sin A + \sin B } = \color{orange}{ 2\sin (\frac{A+B}{2}) \cos (\frac{A-B}{2})}}\]

Answer 27

Now let's put : A = 2 theta and B = theta.

Answer 28

Oh yes thats the one! not the one i was thinking then ha

Answer 29

sum to product formula yes remember now

Answer 30

I made the mistake in previous one . Sorry for that.

Answer 31

its ok any help is great thanks

Answer 32

Now, \[\large{\textbf{we have }}\] :
\[\large{2\sin (2 \theta) \cos \theta = \sin 2\theta}\] Right?

Answer 33

The 2sin(2θ)cosθ being equivalent to sin3θ or all of it. as in being equivalent to sinθ+sin3θ?

Answer 34

^ that is equivalent to \(\large{\sin \theta + \sin 3\theta}\)

Answer 35

yes got you

Answer 36

\[\large{2\sin 2\theta \cos \theta = \sin \theta + \sin 3\theta}\]

and \(\large{\sin \theta + \sin 3\theta = \sin 2\theta}\)

thus : \(\large{\color{blue}{2\sin 2\theta \cos \theta = \sin 2\theta}}\)

Answer 37

Now what do you think should be the next step?

Answer 38

change sin2θ into double angle 2sinθcosθ?

Answer 39

No. See can you divide sin 2 theta both sides?

Answer 40

oh ok so would that leave 2cosθ=1

Answer 41

Yes u can get that (but note if theta can be equal to zero then you can not divide sin 2 theta both sides as 0/0 is indeterminate) .

therefore \(\large{\cos \theta= \frac{1}{2}}\) and also : \(\large{2\sin 2\theta \cos \theta - \sin 2\theta = 0}\)

\(\large{\color{blue}{2\sin 2\theta} (\color{red}{\cos \theta - 1} ) = 0}\)
\(\large{\color{orange}{2\sin 2\theta = 0}}\)

thus sin 2 theta = 9 or cos theta = 1/2

Answer 42

or \(\large{\sin 2\theta (2\cos \theta -1) = 0}\)
=> \(\sin 2\theta = 0\) or \(\cos \theta = \frac{1}{2}\)

got it?

Answer 43

that is fantastic thank you very much. the step by step guide has been great

Answer 44

So did you like : Open + Study ?

Answer 45

yes first time on here and been a big help

Answer 46

Great! So I hope that you will keep asking your doubts here and also do not forget to help others ... :)

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Answer 47

Thanks again i will make sure i will help the level 2 maths. Lot easier

Answer 48

You're welcome and sure that will be great.