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Mathematics 16 Online
OpenStudy (anonymous):

Are there any angles x and y that satisfy sin(x+y)=sinx+siny ? Use an example to explain your answer @kc_kennylau

OpenStudy (kc_kennylau):

Expand the LHS using the formulae hereeeee :) http://www.a-levelmathstutor.com/images/trigonometry/ca-identities01.jpg

OpenStudy (anonymous):

:(

OpenStudy (kc_kennylau):

Just changed the name of the variables, ain't a big deal xD

OpenStudy (anonymous):

sin A + sin B

OpenStudy (kc_kennylau):

can you help me to replace x as A and y as B in the third formula? :)

OpenStudy (anonymous):

I did ^^

OpenStudy (kc_kennylau):

the whole formula...

OpenStudy (kc_kennylau):

\[\sin(A+B)=\sin A\cos B+\cos A\sin B\] Replace the whole formula, A->x and B->y

OpenStudy (anonymous):

Sin(x+y)= sinxcosy+cosxsiny

OpenStudy (kc_kennylau):

exactly

OpenStudy (anonymous):

yay :D

OpenStudy (kc_kennylau):

So can you substitute what you wrote to the left hand side of \(\sin(x+y)=\sin x+\sin y\)? :)

OpenStudy (anonymous):

Idk

OpenStudy (anonymous):

what you mean

OpenStudy (kc_kennylau):

So you have \(\sin(x+y)=\sin x\cos y+\cos x\sin y\) and \(\sin (x+y)=\sin x+\sin y\), what can you say about them? :)

OpenStudy (anonymous):

that they're very similar :P

OpenStudy (kc_kennylau):

|(-_-)|

OpenStudy (kc_kennylau):

If \(a=b\) and \(a=c\), wha can you say about \(b\) and \(c\)?

OpenStudy (anonymous):

its equal to A !!!

OpenStudy (kc_kennylau):

|(-_-)|

OpenStudy (anonymous):

they can equal each other

OpenStudy (anonymous):

:\

OpenStudy (kc_kennylau):

exactlyyyyyyyyyyyyyyyyyyyyyyyy :DDDDDDD

OpenStudy (anonymous):

ohhhh

OpenStudy (anonymous):

so can you summarize what I've just learned ?? thanks

OpenStudy (kc_kennylau):

So if \(\color{blue}{\sin(x+y)}=\color{red}{\sin x\cos y+\cos x\sin y}\) and \(\color{blue}{\sin(x+y)}=\color{green}{\sin x+\sin y}\), what can you say about \(\color{red}{\sin x\cos y+\cos x\sin y}\) and \(\color{green}{\sin x+\sin y}\)?

OpenStudy (kc_kennylau):

you've learnt that when the thing inside \(\sin\) or \(\cos\) is made up of two substances that are the same or different, you can always use the formulae hereeeeeeeeeeeeeeeeeeee xD http://www.a-levelmathstutor.com/images/trigonometry/ca-identities01.jpg

OpenStudy (anonymous):

omg not that link again >.<

OpenStudy (kc_kennylau):

but you have to memorize every. single. formula. of the website.......

OpenStudy (anonymous):

yes yes

OpenStudy (anonymous):

so the example we've used would be ???? what

OpenStudy (kc_kennylau):

would be the first formula used in the previous post, and the third formula being used now :D

OpenStudy (anonymous):

what was the formula from the first tag ??

OpenStudy (kc_kennylau):

what do you mean first taggggg

OpenStudy (anonymous):

first post !

OpenStudy (anonymous):

you said it above

OpenStudy (anonymous):

:'(

OpenStudy (kc_kennylau):

uh, it's the first formula \(\cos(A+B)=\cos A\cos B-\sin A\sin B\)

OpenStudy (anonymous):

ok thanks for putting up with my lack of skils in math

OpenStudy (anonymous):

:)

OpenStudy (kc_kennylau):

no problem :D

OpenStudy (anonymous):

@kc_kennylau oh one more thing !

OpenStudy (anonymous):

how would I convert the angle 4 to degrees???

OpenStudy (kc_kennylau):

\[\fbox{radian}=\fbox{degree}\times\frac\pi{180^\circ}\]\[\fbox{degree}=\fbox{radian}\times\frac{180^\circ}\pi\]

OpenStudy (anonymous):

would it be 1pi/43?

OpenStudy (anonymous):

@kc_kennylau

OpenStudy (kc_kennylau):

was the 4 in radian?

OpenStudy (anonymous):

ikd it says convert following angles to degree measures

OpenStudy (kc_kennylau):

then i assume that it's radian

OpenStudy (anonymous):

so is that the right answer that I got above?

OpenStudy (kc_kennylau):

nope :/

OpenStudy (anonymous):

what is it?

OpenStudy (kc_kennylau):

The formula is \(\fbox{degree}=\fbox{radian}\times\dfrac{180^\circ}\pi\)

OpenStudy (kc_kennylau):

Substitute the \(\fbox{radian}\) as 4 :)

OpenStudy (anonymous):

229.18

OpenStudy (kc_kennylau):

yep :)

OpenStudy (anonymous):

thankss

OpenStudy (anonymous):

:D<3

OpenStudy (kc_kennylau):

no problem <3

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