Tricky trigonometry question

Sin(x+40) +sin(x+50) =0. Solve for the interval (0 to 360)

$\sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos\left( \frac{A-B}{2} \right)$

It's probably in your formula book.

Sin(x+40) +sin(x+50) =0. Solve for the interval (0 to 360)

isn`t this the same as $\sin(2x+ \frac{\pi}{2})$ ??

(you`re is a simpler option, by the way!)

Isn't what the same as $\sin(2x+ \frac{\pi}{2})$?

$\sin(2x+\pi/2)=\cos(2x)$

isn`t this the same as $\sin(2x+ \frac{\pi}{2})$ ??

(you`re is a simpler option, by the way!)

Sin(x+40) +sin(x+50) =0. Solve for the interval (0 to 360)

You can solve this essentially by just thinking about it... You take a value of sin, move ten degrees along and the value is now precisely negative that value so if you were to move only 5 degrees on the value would be what?

Perhaps it is just me but this made no sense whatsoever - could you explain further

You can solve this essentially by just thinking about it... You take a value of sin, move ten degrees along and the value is now precisely negative that value so if you were to move only 5 degrees on the value would be what?

Expanding it is a waste of time but it is useful to know the method.


Posted from TSR Mobile

Sin is anti symmetric about its zeroes.e.g. clearly.


Posted from TSR Mobile

Could you solve the question because I still don't get it

Of course.
$sin(175) + sin(185)=sin(180-5)+sin(180+5)=sin180cos5-sin5cos180+sin180cos5+sin5cos180$

Sin(x+40) +sin(x+50) =0. Solve for the interval (0 to 360)

You can solve this essentially by just thinking about it... You take a value of sin, move ten degrees along and the value is now precisely negative that value so if you were to move only 5 degrees on the value would be what?

Expanding it is a waste of time but it is useful to know the method.


Posted from TSR Mobile

Our of curiosity, why did the below fail?

$sin(x+40)=-sin(x+50)=sin(-x-50)$
$x+40=-x-50$
$2x=-90$
$x=-45$
$x=315$