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Express 3cosx - 4sinx as rcos(x+Φ)

Hi, there is a C3 question which I can understand mostly, apart from getting the 'r'.

In my revision guide it has:

rcos(x+Φ) ≡ 3cosx-4sinx

--> rcosxcosΦ-rsinxsinΦ ≡ 3cosx-4sinx
--> rcosΦ = 3, rsinΦ = 4

I understand all that. But the next line is:

--> r = √(3²+4²) = 5

Where the hell does that come from? :confused:

It's probably really simple but I just can't get it! Could anybody explain this to me?

Reply 1

RichE

RhiddynUK

Hi, there is a C3 question which I can understand mostly, apart from getting the 'r'.

In my revision guide it has:

rcos(x+Φ) ≡ 3cosx-4sinx

--> rcosxcosΦ-rsinxsinΦ ≡ 3cosx-4sinx
--> rcosΦ = 3, rsinΦ = 4 (*)

I understand all that. But the next line is:

--> r = √(3²+4²) = 5

Where the hell does that come from? :confused:

It's probably really simple but I just can't get it! Could anybody explain this to me?

Square equations (*) and add

Reply 2

meathead

Remember that cos²Φ + sin²Φ ≡ 1

Therefore
r² = r²(cos²Φ + sin²Φ ) = r²cos²Φ + r²sin²Φ = (r cosΦ )² + (r sinΦ )² = 3² + 4²

so r = √(3² + 4²)

Or another way to look at it:

r cos Φ = adjacent side of a triangle with hypotenuse r
r sin Φ = opposite side of a triangle with hypotenuse r

adjacent is 3, opposite is 4, therefore r = hypotenuse = √(3²+4²) = 5

Reply 3

Exile

rcosx = 3 (1)
rsinx = 4 (2)

(1)² +(2)²

r²cos²x+rsin²x=3²+4²
r²(cos²x+sin²x)=25 (take common factor r² out)
r²(1)=25 (identity cos²x+sin²x=1)
r²=25
r=5

Reply 4

ChanandlerBong