Question about Auxillary Method - Trig (1 Viewer)

[Smilebuffalo]

When looking at past HSC solutions the answer always derives how to transform asinx + bcosx into Rsin(x + $)

I was wondering if we have to derive it as well, or can we simply use the formula for finding R and $ using:

R = sqrt(a^2 + b^2) and tan $ = b/a


???

 

[addikaye03]

When looking at past HSC solutions the answer always derives how to transform asinx + bcosx into Rsin(x + $)

I was wondering if we have to derive it as well, or can we simply use the formula for finding R and $ using:

R = sqrt(a^2 + b^2) and tan $ = b/a


???

Just have to use the formula i believe

 

[Michaelmoo]

You can use the formula. Don't have to.

 

[biopia]

I find it better to derive it as you go.
For example: Express sinθ +√3cosθ in the form of Rsin(θ-α)

First, Rsin(θ-α) = Rsinθcosα - Rcosθsinα

Equating gives;
sinθ = Rsinθcosα ⇒ 1 = Rcosα [divide both sides by sinθ]
Similarly,
√3cosθ = Rcosθsinα ⇒ √3 = Rsinα

So we have:
Rsinα = √3
Rcosα = 1
By division, it is clear that,
tanα = √3
α = π/3

Use the formula part for R
1² + √3² = R² ⇒ R = 2

Therefore:
sinθ +√3cosθ = 2sin(θ-π/3)

I have always done it this way. It shows the markers you aren't just wrote learning.
I suppose if you use the formula, it's ok too. I just thought I'd contribute the derivation.
=]

 

[Prashant Sallan]

just learn the formula, you have 1000 other things to remember, biopia - you machine.

 

[madsam]

Express sinθ +√3cosθ in the form of Rsin(θ+@)
--> 2(1/2 sinθ + √3/2cosθ)
I find it easier to simply force in an R value

then
1/2 = cos@

√3/2 = sin@

Draw the triangle, and you get @ = pi/3

--> 2sin(θ+pi/3)

 

[biopia]

biopia - you machine.

Haha, am I right to take that as a compliment?
Thanks lol.