help the dumb dumb again (1 Viewer)

[me dumb]

i)express sin4t + sqrt3(cos4t) in the form Rsin(4t +a) where a is in radians
ii)hence solve sin4t + sqrt3(cos4t)=0
part ii) i could do without the first proof, but i would still like to know how to do it with the above proof. mainly becasue the way i did it was easy
divide both sides by cos4t
tan4t=-sqrt3
easy =)
any help is appreciated

 

[danal353]

you're not dumb... i don't even do extension maths...

 

[me dumb]

you're not dumb... i don't even do extension maths...

hopefully it attracts peoples attention
hehe well ive never seen the above type question be4 and i only have simple harmonic motion left, so i missed something
what topic does this go under?

 

[danal353]

trigonometry? when it comes to maths, i'm really not the right person to ask...

 

[addikaye03]

i)express sin4t + sqrt3(cos4t) in the form Rsin(4t +a) where a is in radians
ii)hence solve sin4t + sqrt3(cos4t)=0
part ii) i could do without the first proof, but i would still like to know how to do it with the above proof. mainly becasue the way i did it was easy
divide both sides by cos4t
tan4t=-sqrt3
easy =)
any help is appreciated

i) ok, So R=rt (a^2+b^2) so R=rt ( 1^3 +(sqrt 3)^2)= rt4=2

tan(b/a)=rt (3), by looking at exact value triangles:

=pi/3

Thereofore sin4t + sqrt3(cos4t)=2sin(4t+pi/3)

ii) Now we set the above equal to 0, 2sin(4t+pi/3)=0

4t+pi/3=0; pi where 0<=x<=2pi

therefore t= -pi/12, pi/6 #

Alternative:

ii)sin4t + sqrt3(cos4t)=0

sin^2(4t)+3cos^2(4t)=0

sin^2(4t)+3(1-sin^2(4t))=0

sin^2(4t)+3-3sin^2(4t)=0

2sin^2(4t)-3=0...then solve accordingly

But since it says "hence" this CANNOT be done, if it said "Hence or otherwise" then go ahead

 

[Pwnage101]

Basically, the theory behind this 'auxiliary angle method' is that, by expoansion,

Rsin(4t +a)=R[sin(4t)cos(a)+sin(a)cos(4t)].......... (1)

we have sin4t + sqrt3(cos4t)

know take the coefficient of sin4t and cos4t and square root the sum of their squares to get R

ie if it si Asin4t + Bcos4t, R would equal root(A^2+B^2)

so in this case R = 2

so we can write sin4t + sqrt3(cos4t)

as 2[(1/2)*sin(4t)+(sqrt3/2)cos(4t)] ....(2)

now we equate and compare (1) and (2), put one on the line under the other.

R[sin(4t)cos(a)+sin(a)cos(4t)]

=

2[(1/2)*sin(4t)+(sqrt3/2)cos(4t)]

so R = 2

cos(a) = (1/2) and sin(a) = (sqrt3/2) by comparing the coefficients in each

so it is clear, in radians, a = pi/3

thus sin4t + sqrt3(cos4t) = 2sin(4t+pi/3)

addikaye03 did (ii)

 

[me dumb]

i) ok, So R=rt (a^2+b^2) so R=rt ( 1^3 +(sqrt 3)^2)= rt4=2
quote]
sorry, why does R=rt(a^2+b^2) and why does 1^3=a^2 and (rt3)^2=b^2
wat i assume it is is the number infront of the sin and cos?
thnx

 

[me dumb]

Basically, the theory behind this 'auxiliary angle method' is that, by expoansion,

Rsin(4t +a)=R[sin(4t)cos(a)+sin(a)cos(4t)].......... (1)

we have sin4t + sqrt3(cos4t)

know take the coefficient of sin4t and cos4t and square root the sum of their squares to get R

ie if it si Asin4t + Bcos4t, R would equal root(A^2+B^2)

so in this case R = 2

so we can write sin4t + sqrt3(cos4t)

as 2[(1/2)*sin(4t)+(sqrt3/2)cos(4t)] ....(2)

now we equate and compare (1) and (2), put one on the line under the other.

R[sin(4t)cos(a)+sin(a)cos(4t)]

=

2[(1/2)*sin(4t)+(sqrt3/2)cos(4t)]

so R = 2

cos(a) = (1/2) and sin(a) = (sqrt3/2) by comparing the coefficients in each

so it is clear, in radians, a = pi/3

thus sin4t + sqrt3(cos4t) = 2sin(4t+pi/3)

addikaye03 did (ii)

o i undertand now thanyou