GCSE maths help

[Jennaveeve]

Can anyone tell me if this is right? DS has the past paper but not the answers and I can’t find them online - and frankly, my maths can’t cope with this level of question 😂

Compare this curve to the cosine and the sine curves and you should see that it’s most similar to the cosine curve that starts and the value of y=1 when x=0 (just put cos(0) into a calculator and you get the result of 1). Normally cosine curve will start at y=1 at x=0 and drop down to y=0 at x=90 then drop further down to y=-1 at x=180 then rise back up to y=0 at x=270 then back up to y=1 again, having completed one cycle, at x=360 degrees. Now the curve shown in the question differs in several ways - instead of the normal oscillation from y=-1 to y=+1, a range of 2 from the minimum of y=-1 up to the maximum of y=+1, your curve goes from a minimum of y=-1 to a maximum of y=+3, a range of 4 - it is stretched or magnified in the y direction by a factor of 2 compared to a normal cosine wave - the magnifying factor is the value of variable “a” so a=2.

Your wave completes 3 complete oscillations (an oscillation starts from y=1, goes down to y=-1 and back up to y=+1 on a normal cosine curve) between x=0 and x=360 - this factor of 3 is the value of the variable “b” - we tend to talk in terms of variable b being the frequency multiplier.

Now a normal cosine wave is centred around the x axis - it’s symmetrical around the x axis - it goes up to y=+1 and down to y=-1 but the curve in the question is not centred or symmetrical around the x axis - it is shifted up in the y axis. It is also stretched by a factor of 2 in the y direction but it is also shifted up by 1 unit in the y direction - this shift in the y direction is represented by the variable c so c=1.

a = (Maximum - Minimum)/2 = (3 - -1)/2 = 2

(If the cosine graph had been reflected in the x-axis, a would be -2)

b = the number of cycles in 360° = 3

c = (Maximum + Minimum)/2 = (3 + -1)/2 = 1