Competition - post a photo you've taken of nature >>

A-level Trig Help

It is given that f(x)=sin(x+30)+cos(x+60)
a) show that f(x)=cosx, hence show that f(4x)+4f(2x) = Acos^4x-B
b) determine the greatest and least values of 1/f(4x)+4f(2x)+7 as x varies
c) solve the equation, sin(12a+30)+cos(12a+60)+4sin(6a+30)+4cos(6a+60)=1 for 0<a<60

I have worked out part a as A=8 and B=3, which is correct, however I am struggling to find a starting point for both parts b and c, any ideas?

Reply 1

mathstutor24

Think about how you would use your answer to part a) in the equation stated in part b)

Reply 2

alicejanes

Original post by mathstutor24

Think about how you would use your answer to part a) in the equation stated in part b)

I have now got 1/(8cos^4x + 4) which I have written in terms of sec, 8sec^4x + 1/4. However there is no equals so i know i cannot move the 1/4 across, so I dont know where to go from this equation.

Reply 3

mathstutor24

Think about the range of y=cos(x)

i.e. what is the biggest and smallest value cos(x) can equal. How can you use this information to determine the greatest and least values?

(edited 3 years ago)

Reply 4

alicejanes

Original post by mathstutor24

Think about the range of y=cos(x)

I'm sorry, I dont understand

Reply 5

mathstutor24

what is the biggest value that cos(x) can be, regardless of the value of x?
similarly, what is the smallest value that cos(x) can be?

Annotation 2020-06-08 171015.png

Reply 6

alicejanes

Original post by mathstutor24

what is the biggest value that cos(x) can be, regardless of the value of x?
similarly, what is the smallest value that cos(x) can be?

Annotation 2020-06-08 171015.png

so 1 and -1, would I make 1/8cos^4x+4 equal to each of these values and then solve ?

Reply 7

mathstutor24

Almost! Yes, the largest and smallest values of cos(x) are 1 and -1. So... if you think about them in terms of

cos^{4}x

they will both be 1, so setting cos(x) to 1 and -1 will give you the same answer. HINT: consider cos(x)=0

Reply 8

davros

Original post by alicejanes

I have now got 1/(8cos^4x + 4) which I have written in terms of sec, 8sec^4x + 1/4.

Are you sure those 2 expressions are equivalent?

Hint: you don't actually need to "convert" the fraction - you're just interested in how big or small the denominator can be so you can work out how small (or big) the overall fraction can be :smile:

Reply 9

mathstutor24

cos(x)=1

, then

cos^{4}(x)=1^{4}=1

Similarly, if

cos(x)=-1

, then

cos^{4}(x)=(-1)^{4}=1

Reply 10

mathstutor24

Original post by davros

That's what I'm trying to help them see, @davros :smile:

Reply 11

davros

Original post by mathstutor24

That's what I'm trying to help them see, @davros :smile:

Yes I think we were typing at the same time - I was concerned they were going down the wrong route with that invalid fraction conversion and weren't following your excellent advice :smile:

Reply 12

mathstutor24

The bit that I've highlighted in blue - you should be able to figure out the max and min value that can equal using the info I've posted already. As @davros said, you need to calculate the biggest and smallest denominator, which you can do from your knowledge of the range of cos(x).

Reply 13

alicejanes

Original post by mathstutor24

cos(x)=1

, then

cos^{4}(x)=1^{4}=1

Similarly, if

cos(x)=-1

, then

cos^{4}(x)=(-1)^{4}=1

so:
1/(8x1)+4 = 1/12 would be one of the values and the other would be 1/(8x0)+4 = 1/4 ??

Reply 14

mathstutor24

Yes

This kind of thing (where they ask you for a min or max value involving trig) comes up frequently in A Level maths. It's usually involving sine and cosine, so remember a tip is to remember your trig graphs - the max and min values that sin(x) and cos(x) can be - and figure it out from there. If your equation didn't include an even exponent, then you would have used the min value of cos(x) (i.e. -1) to get the least value, since (-1)^odd would result in a negative value.