Auxiliary method (2 Viewers)

[fifibum]

When solving a question like 3sintheta + 4costheta = 0,
how would you know which sum or difference of two angles to use?
For example, what makes you choose to use sin(theta + alpha)
instead of cos(theta + alpha)?

 

[InteGrand]

 

[fifibum]

How would you actually do this question though?

 

[leehuan]

?



Then just write down the general solution or the required answers for your unspecified domain?

______________________


Or alternatively you can just solve this?

 

[Green Yoda]

ok for
sinx+cosx use rsin(x+alpha)
sinx-cosx use rsin(x-alpha)
cosx+sinx use rcos(x-alpha)
cosx-sinx use rcos(x+alpha)

 

[fluffchuck]

3sintheta + 4costheta = 0

Let 3sintheta + 4costheta = Rcos(theta-alpha)
= Rcostheta.cosalpha+Rsintheta.sinalpha

By comparison, 3sintheta = Rsintheta.sinalpha
3 = Rsinalpha

By comparison, 4costheta = Rcostheta.cosalpha
4 = Rcosalpha

Therefore:
Rsinalpha = 3------Eq1 (Quadrants 1,2)
Rcosalpha = 4------Eq2 (Quadants 1,4)
Hence, Quadrant 1

(Eq1)^2 + (Eq2)^2=
R^2(sin^2alpha+cos^2alpha)= 3^2 + 4^2
Since sin^2alpha+cos^2alpha = 1,
R^2 = 25
R=5 (R>0)

Rsinalpha = 3------Eq1 (Quadrants 1,2)
Rcosalpha = 4------Eq2 (Quadants 1,4)

Eq1 divided by Eq2
tanalpha = 3/4
alpha = tan^-1(3/4) (Take the acute angle, since quadrant 1)

Therefore,
3sintheta + 4costheta = 5cos(theta - tan^-1(3/4))

Since 3sintheta + 4costheta = 0, 5cos(theta - tan^-1(3/4)) = 0

Solve for 5cos(theta - tan^-1(3/4)) =0.

5cos(theta - tan^-1(3/4)) = 0
cos(theta - tan^-1(3/4)) = 0
theta = 180n + tan^-1(3/4), where n is an integer

Then, substitue values for n for angles between your questions given condition

Hope this helped

 

[InteGrand]

3sintheta + 4costheta = 0

Let 3sintheta + 4costheta = Rcos(theta-alpha)
= Rcostheta.cosalpha+Rsintheta.sinalpha

By comparison, 3sintheta = Rsintheta.sinalpha
3 = Rsinalpha

By comparison, 4costheta = Rcostheta.cosalpha
4 = Rcosalpha

Therefore:
Rsinalpha = 3------Eq1 (Quadrants 1,2)
Rcosalpha = 4------Eq2 (Quadants 1,4)
Hence, Quadrant 1

(Eq1)^2 + (Eq2)^2=
R^2(sin^2alpha+cos^2alpha)= 3^2 + 4^2
Since sin^2alpha+cos^2alpha = 1,
R^2 = 25
R=5 (R>0)

Rsinalpha = 3------Eq1 (Quadrants 1,2)
Rcosalpha = 4------Eq2 (Quadants 1,4)

Eq1 divided by Eq2
tanalpha = 3/4
alpha = tan^-1(3/4) (Take the acute angle, since quadrant 1)

Therefore,
3sintheta + 4costheta = 5cos(theta - tan^-1(3/4))

Since 3sintheta + 4costheta = 0, 5cos(theta - tan^-1(3/4)) = 0

Solve for 5cos(theta - tan^-1(3/4)) =0.

5cos(theta - tan^-1(3/4)) = 0
cos(theta - tan^-1(3/4)) = 0
theta = 180n + tan^-1(3/4), where n is an integer

Then, substitue values for n for angles between your questions given condition

Hope this helped

You accidentally wrote the general solution for sin rather than cos (when you did the 180n degrees at the end, that's the general solution to sin X = 0 rather than cos, which is what we needed).

 

[fluffchuck]

You accidentally wrote the general solution for sin rather than cos (when you did the 180n degrees at the end, that's the general solution to sin X = 0 rather than cos, which is what we needed).

Oh yes, my apologies,

Resuming from: Solve for 5cos(theta - tan^-1(3/4)) =0.

Solve for 5cos(theta - tan^-1(3/4)) =0.

5cos(theta - tan^-1(3/4)) = 0

cos(theta - tan^-1(3/4)) = 0

theta = 360n +/- tan^-1(3/4), where n is an integer

 

[fifibum]

Thanks heaps!