Help: trig equations

[synopsys]

Could someone tell me how to solve this equations ? I've spent like 2 days:

sin(x)*sin(10) = sin(20)*sin(250-x)

Thanks.

 

[mmm4444bot]

Welcome to the tutoring boards. Please post some of your work or explain what you're thinking.

Are you stuck somewhere specific? Perhaps, you were on the right track, but made a simple mistake.

Tutors require basic information about your situation in order to guide you.

Cheers :cool:

 

[synopsys]

Welcome to the tutoring boards. Please post some of your work or explain what you're thinking.

Are you stuck somewhere specific? Perhaps, you were on the right track, but made a simple mistake.

Tutors require basic information about your situation in order to guide you.

Cheers :cool:

I am not sure I am on the right track:

sin(x)*sin(10)=2*sin(10)*cos(10)*sin(240 + (10-x) )


sin(x)=2*cos(10)*(sin(240)*cos(10-x) + cos(240)*sin(10-x))


= -cos(10)*(sqrt(3)*cos(10)*cos(x)+sqrt(3)*sin(10)*sin(x)+sin(10)*cos(x)-cos(10)*sin(x))




cos(x)*(sqrt(3)*cos(10)*cos(10)+sin(10)*sin(10)) + sin(x)*(1 + sqrt(3)*sin(10)*cos(10) -cos(10)*cos(10))=0;


tan(x)= - (sqrt(3)*cos(10)*cos(10)) / ( 1 + sqrt(3)*sin(10)*cos(10) - cos(10)*cos(10)) ;


cannot go any further...

 

[soroban]

Hello, synopsys!

\(\displaystyle \text{Solve for }x\!:\;\sin x\sin10 \:=\: \sin20\sin(250-x)\)

I did it like this . . .


. . . . . . . . . . . . . . . . . .\(\displaystyle \sin10\sin x \:=\:\sin20(\sin250\cos x - \cos250\sin x)\)

. . . . . . . . . . . . . . . . . .\(\displaystyle \sin10\sin x \:=\:\sin20\sin250\cos x - \sin20\cos250\sin x\)

. . \(\displaystyle \sin10\sin x + \sin20\cos250\sin x \:=\:\sin20\sin250\cos x\)

. . . . .\(\displaystyle (\sin10 + \sin20\cos250)\sin x \:=\:\sin20\sin250\cos x\)

. . . . . . . . . . . . . . . . . . . . . \(\displaystyle \dfrac{\sin x}{\cos x} \:=\:\dfrac{\sin20\sin250}{\sin10 + \sin20\cos250}\)

. . . . . . . . . . . . . . . . . . . . . \(\displaystyle \tan x \;=\;\dfrac{\sin20\sin250}{\sin10 + \sin20\cos250}\)

. . . . . . . . . . . . . . . . . . . . . . . .\(\displaystyle x \;=\;\tan^{\text{-}1}\!\left(\dfrac{\sin20\sin250}{\sin10 + \sin20\cos250}\right) \)

 

[synopsys]

Hi, soroban,

Thanks a lot for your reply.
However, this is a contest problem and no calculator can be used.
I used calculator and found the answer should be 100. (x is an angle of a triangle).
Is there any way x can be found by simplification ?

Thanks.

 

[Jameson]

I haven't tried, but I think using the product to sum identities could be useful here since a calculator isn't allowed. Also, sin20 can't be expressed rationally so cancellation is going to happen. Try rewriting the products as sums.

 

[synopsys]

I tried but still couldn't solve the problem.

 

[Deleted member 4993]

Hi, soroban,

Thanks a lot for your reply.
However, this is a contest problem and no calculator can be used.


Thanks.

You ARE allowed to seek external help in this contest?

 

[synopsys]

You ARE allowed to seek external help in this contest?

I am not competing in this contest. It ended many years ago.