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Maths

fθ=5cosθ+12sinθ in the form R(cosθ-α) is 13cos(θ-67.4)
using the equation 2/13cos(θ-67.4)

How can I find the maximum value? I know its Max/Min

Reply 1

Dynamic_Vicz

The maximum occurs where cos(theta-67.4)=1. What value of theta does this occur for ?

(edited 6 years ago)

Reply 2

username2053059

Original post by joyoustele

fθ=5cosθ+12sinθ in the form R(cosθ-α) is 13cos(θ-67.4)
using the equation 2/13cos(θ-67.4)

How can I find the maximum value? I know its Max/Min

What is this madness?

Reply 3

joyoustele

Original post by Dynamic_Vicz

The maximum occurs where cos(theta-67.4)=1. What value of theta does this occur for ?

Max value when theta =67.4

Reply 4

joyoustele

Never mind, Classic me, i read the question wrong. It's 30/13cos(θ-67.4) not 2/13cos(θ-67.4)

Reply 5

Dynamic_Vicz

Original post by joyoustele

Max value when theta =67.4

Correct so you know that cos(theta - 67.4) =1 is the maximum value when theta =67.4.

All you need to do is multiply 1 by 13 because you're trying to find the maximum value of 13cos(theta-67.4)

Reply 6

joyoustele

Original post by Dynamic_Vicz

Correct so you know that cos(theta - 67.4) =1 is the maximum value when theta =67.4.

All you need to do is multiply 1 by 13 because you're trying to find the maximum value of 13cos(theta-67.4)

I did all that but didnt get the correct answer, but came to realise i read the question wrong.
it was actually the maximum of 30/13cos(θ-67.4)+2 which i believe is 2. As 30/13*1+2 =2 correct?

Reply 7

Dynamic_Vicz

Original post by joyoustele

I did all that but didnt get the correct answer, but came to realise i read the question wrong.
it was actually the maximum of 30/13cos(θ-67.4)+2 which i believe is 2. As 30/13*1+2 =2 correct?

If you saying 30/(13cos(θ-67.4)+2) then the maximum value isn't 2. This would be the minimum value.

You'd want the denominator to be as small as possible. So basically 13cos(θ-67.4) =-1 so that you get 30/(-1+2)=30/1=30