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# A2 c3 help please watch

1. Q: express 3sin+4cos in the form Rsin(theata + alpha). And 0< alpha < 90 degrees
(!)= theata
(?)= alpha

ii) hence, solve the equation 3sin(!) + 4cos(!) +1 = 0

solutions between -180 and 180

Ive done part1 which gives the angle 53.1.....

the he second part I'm stuck with as I don't think I'm correct, here is my working out for part 2

3sin(!) +4cos(!) = inverse tan(4/3)

therefore re can I just make it

tan = 4/3 -1

inverse tan = 18.4349.....

i have no clue what I'm doing thus me guessing what to do...
Q: express 3sin+4cos in the form Rsin(theata + alpha). And 0< alpha < 90 degrees
(!)= theata
(?)= alpha

ii) hence, solve the equation 3sin(!) + 4cos(!) +1 = 0

solutions between -180 and 180

Ive done part1 which gives the angle 53.1.....

the he second part I'm stuck with as I don't think I'm correct, here is my working out for part 2

3sin(!) +4cos(!) = inverse tan(4/3)

therefore re can I just make it

tan = 4/3 -1

inverse tan = 18.4349.....

i have no clue what I'm doing thus me guessing what to do...
have you got any notes from school?
3. Yes, just flicking through them
Q: express 3sin+4cos in the form Rsin(theata + alpha). And 0< alpha < 90 degrees
(!)= theata
(?)= alpha

ii) hence, solve the equation 3sin(!) + 4cos(!) +1 = 0

solutions between -180 and 180

Ive done part1 which gives the angle 53.1.....

the he second part I'm stuck with as I don't think I'm correct, here is my working out for part 2

3sin(!) +4cos(!) = inverse tan(4/3)

therefore re can I just make it

tan = 4/3 -1

inverse tan = 18.4349.....

i have no clue what I'm doing thus me guessing what to do...
Try using your answer from part i) to solve part ii) so..

Rsin(θ + 53.1) = -1 (using the value you calculated for R)
5. So it would be

inverse tan-1/5 - 53.1.... Then makes missing theata number ?
So it would be

inverse tan-1/5 - 53.1.... Then makes missing theata number ?
Why do you keep doing "inverse tan"?

You have an equation that is basically

sin(something) = something else

so the first thing you should be doing is an inverse sine!
7. Sin(theata +53.1) = -1/5

can I then inverse sin to make

theata + 53.1 = inverse sin(-1/5)

then - 53.1

theata = -64.63...

is is that how you do it ?
Sin(theata +53.1) = -1/5

can I then inverse sin to make

theata + 53.1 = inverse sin(-1/5)

then - 53.1

theata = -64.63...

is is that how you do it ?
Eseentially, although you'll need to do something to get angles within the range of values they've asked for.

(and it's theta by the way )
9. Yh I got the rest thanks, I found the angles between the range.

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Updated: December 6, 2014
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