Area between two curves in a set limit - both curves below x-axis - what should I do?

this is the question. part a is simple.

part b - rubbish.

I know we should use sin2x=2sinXcosX,

but if tan = sin/cos = 3/4 why do they get the answer sinX= 3/5 and cosX = 4/5

I know this is possible, but how do they arrive to this answer, where does the 5 come from? It could easily be 3/10 or 3/7 for all I know.

Anyone explain/help/solve?

I have the answer only from the past paper, but haven't got a clue on how to do it.
thanks.

i am not sure exactly whats the question, but if the question is

if tan = sin/cos = 3/4 why do they get the answer sinX= 3/5 and cosX = 4/5


then, think of a right angled triangle.



tan A = 3/4
do you know SOH CAH TOA? (sine=opposite/hypothenuse,cos=adjacent/hypothenuse, tan=opposite/adjacent)... opposite = length of line opposite to the angle. hypothenuse length = opposite right angle (longest length). adjacent=length of line that is at the angle x and 90degrees (if you know what i mean)
well tan A = opposite lenght/adjacent length = a/b
sincce tan A = a/b = 3/4 ... A=3, B=4.
using pythangoras you can find C = sq root (3^2+4^2) = 5
now sin A = opposite length/hypothenuse length = a/c = 3/5
and cos A = adjacent length/hypothenuse length = b/c = 4/5

hope you understand :/

aaaaaah, well I thought about the triangle being right-angled, but it's not mentioned in the question. So we were meant to deduce it from sin and cos - I know that we use cos and sin not only in right-angled triangles, so I got confused. sin and cos as relationship between sides is for right-angled triangles only, but we can use sine and cosine rule with any triangles, right?

Thanks a lot for the explanation.

yup, sine and cosine rules can be used with any triangles, but when you know sin x = a/b or somefin, you can use the right angle triangle to find cos x and tan x, and vice versa..

yup, sine and cosine rules can be used with any triangles, but when you know sin x = a/b or somefin, you can use the right angle triangle to find cos x and tan x, and vice versa..

If we have 3^(log34) does it just equal 4?

So is there a general formula x^(logxY) = Y ?

Thanks

break up the function:

4cos((3X+2)/6) - 1

the 4 doesnt affect period, just the maximum and minimum values
the -1 doesnt affect period either, it just shifts the graph down one

looking inside the cos, youve got:
cos( (3x+2)/6 )
this could be written as cos(3x/6 + 2/6)

does that help?

i've got another question, anyone?

Just state what it is!

it's at the top, sorry.

I'm trying to avoid spamming maths forum with new threads every time I get stuck on some question.

My maths exam is in a week and I'm rubbish at maths. I'm glad people help me out here

You don't get where the limits came from?

Looks like they forgot to include them in the original question.

Remember that the integral of sin x is - cos x

and when you integrate sin (2x-3), you have to differentiate the bracket to get 2 and divide the integral by it thus you get -1/2 cos (2x-3) + C

I think you're confused about the 2x-3 being a function of the sin... imagine y=2x-3, then the question is asking for the integral of sin y, not of sin (2x) - 3...