Year 12s: what will be the first way you research your future options? >>

C3 Trig

Express f(θ) = 5 cos θ + 12 sin θ in the form RCos (θ-α)
I got: 13 cos (θ - 1.176)

part b) Find the smallest possible value of

\frac{30}{ f(\theta) + 2}

I'm completely stuck on how to do the second part, for the smallest value it would be -1 not sure how this would help me get the answer, thanks for all the help!

Scroll to see replies

Reply 1

TenOfThem

Original post by IgorYakov

Express f(θ) = 5 cos θ + 12 sin θ in the form RCos (θ-α)
I got: 13 cos (θ - 1.176)

part b) Find the smallest possible value of

\frac{30}{ f(\theta) + 2}

I'm completely stuck on how to do the second part, for the smallest value it would be -1 not sure how this would help me get the answer, thanks for all the help!

What does this mean

Reply 2

IgorYakov

Original post by TenOfThem

What does this mean

Sorry I didn't word that properly, for the smallest value for Cos it would be -1 right? So in the case of the 13 cos (θ - 1.176) it would be -13?

Or is this completely different to maximum and minimum values?

Reply 3

user2020user

Original post by IgorYakov

Express f(θ) = 5 cos θ + 12 sin θ in the form RCos (θ-α)
I got: 13 cos (θ - 1.176)

part b) Find the smallest possible value of

\frac{30}{ f(\theta) + 2}

I'm completely stuck on how to do the second part, for the smallest value it would be -1 not sure how this would help me get the answer, thanks for all the help!

f(\theta) = R\cos \left( \theta - \alpha \right), \ \text{where} \ \alpha = \text{arctan} \left( \frac{12}{5} \right) \ \text{and} \ R=13

Which value of

f(\theta)

would force

\frac{30}{ f(\theta) + 2}

to take on it's smallest value?

Hint

Further Hint

(edited 10 years ago)

Reply 4

TenOfThem

Original post by IgorYakov

Sorry I didn't word that properly, for the smallest value for Cos it would be -1 right? So in the case of the 13 cos (θ - 1.176) it would be -13?

Your smallest value of f(theta) = -13

So what are the smallest and largest values of f(theta) + 2

Reply 5

IgorYakov

Original post by Khallil

f(\theta) = R\cos \left( \theta - \alpha \right), \ \text{where} \ \alpha = \text{arctan} \left( \frac{12}{5} \right) \ \text{and} \ R=13

Which value of

f(\theta)

would force

\frac{30}{ f(\theta) + 2}

to take on it's smallest value?

Hint

Further Hint

Ohh okay so for the maximum of 13 cos (θ - 1.176) it is 13 as the maximum for a cos graph is 1 :biggrin:

Reply 6

IgorYakov

Original post by TenOfThem

Your smallest value of f(theta) = -13

So what are the smallest and largest values of f(theta) + 2

Wouldn't you just have one value of -13+2 for the minimum of -13?

Reply 7

TenOfThem

Original post by IgorYakov

Wouldn't you just have one value of -13+2 for the minimum of -13?

-11 is indeed the minimum

what is the maximum

Reply 8

user2020user

Original post by IgorYakov

Ohh okay so for the maximum of 13 cos (θ - 1.176) it is 13 as the maximum for a cos graph is 1 :biggrin:

Now what does the fact that the maximum value of

f(\theta)

being 13 tell you about

\frac{30}{ f(\theta) + 2} \ ?

Reply 9

IgorYakov

Original post by Khallil

Now what does the fact that the maximum value of

f(\theta)

being 13 tell you about

\frac{30}{ f(\theta) + 2} \ ?

Original post by TenOfThem

-11 is indeed the minimum

what is the maximum

That it's 2 :biggrin:

Thanks! I was trying to find the minimum and not the maximum, I didn't realise that because it's on the denominator I needed the maximum for the whole number to be it's smallest

Khallil PRSOM

Reply 10

TenOfThem

Original post by Khallil

f(\theta) = R\cos \left( \theta - \alpha \right), \ \text{where} \ \alpha = \text{arctan} \left( \frac{12}{5} \right) \ \text{and} \ R=13

Which value of

f(\theta)

would force

\frac{30}{ f(\theta) + 2}

to take on it's smallest value?

Hint

Further Hint

You need to look at the specific question as your hint is not correct in this case

Reply 11

TenOfThem

Original post by IgorYakov

That it's 2 :biggrin:

Thanks! I was trying to find the minimum and not the maximum, I didn't realise that because it's on the denominator I needed the maximum for the whole number to be it's smallest