A maths - Trigo Graphs

bonkysleuth SG

Hey there, here's an equation which I have difficulty drawing out the graph.

|4 cos x/2 - 1 | for 0 < or equal x < or equal to pi

amplitude's 4 and period is 4 pi.

I drew out the graph but it looks weird. Would like someone to help me with this. Thank you. (:

And yea, I wanna check the shape of the graph for y =|2 tan 3x|

^tamago^ SG

bonkysleuth SG

Thanks ^tamago^

There is this question which I have problem solving. I got the wrong answers although I think my graph has got no problem. Can someone please take a look?

Sketch, on the same diagram, the graphs of y = sin x and y = 2 cos x, for the values of x from 0 and 2 pi. Hence,state

(a)the number of roots of the equation sin x = 2 cos x in the range 0 to 2 pi

(b) the range of values of x, between 0 and 2 pi, for which sin x and 2 cos x are both decreasing as x increases

For (a) the actual answer is 2... but i got 4 instead.

As for (b), my range is pi<x<3pi/2. the answer sheet writes pi/2 <x <pi

If possible, can someone help check whether I have done wrongly or are the answers correct?

Another question goes like this.

Sketch, on the same diagram, the graphs of y =3 cos x -2 and y =4 |sin x| for the doman 0 deg< or equal to x< or equal to 360 deg. Hence deduce the value of k for which the equation 3 cos x -2 = |4 sin x| + k has 3 solutions in this domain.

Is it really possible to solve this graph? I realise all the degree units are too close to one another such that one cannot even really read the values correctly.

Thanks.

^tamago^ SG

eagle SG

wah... tamago... which software you used?

^tamago^ SG

Mac OS X’s Grapher. (:

^tamago^ SG

k = -5 btw. (:

bonkysleuth SG

For the above question where |4 cos x/2 - 1 | for 0 < or equal x < or equal to pi, the axis starts at -1. So do I start reflecting my graph upwards after it passes the -1 axis or at 0?

^tamago^ SG

the whole thing is modulus, so as long as y < 0, reflect.