[zincoff]

(Original post by raheem94)


Take the cos of both sides



We know

Hence,

Thanks .
Yup I understood that section, but what I was asking was why using sin(y+π/2) on a graph transformation would not give the correct answer. Sorry if I didn't phrase that clearly.

[arvin_infinity]

(Original post by raheem94)
You both are wrong, see my post.

That is so weird ..at 1:25 I edited...at 1:25 you posted this --


em how can we differentiate

[raheem94]

(Original post by arvin_infinity)
That is so weird ..at 1:25 I edited...at 1:25 you posted this --


em how can we differentiate

May be we both did it at the same time.

For these questions use the chain rule.



I will do one question for you.



Differentiating it gives,

[arvin_infinity]

(Original post by raheem94)
May be we both did it at the same time.

For these questions use the chain rule.



I will do one question for you.



Differentiating it gives,

I was meant to ask a totally different question..was gonna say what is the integration of those- might as well ask them later on for C4 page then

[raheem94]

(Original post by zincoff)
Thanks .
Yup I understood that section, but what I was asking was why using sin(y+π/2) on a graph transformation would not give the correct answer. Sorry if I didn't phrase that clearly.

We usually write it as

I can't think of the answer to your question right now.

The results are usually quoted as 90-x in the books.

[raheem94]

(Original post by arvin_infinity)
I was meant to ask a totally different question..was gonna say what is the integration of those- might as well ask them later on for C4 page then

To integrate we differentiate

Hence,

Others will be integrated in a similar way.

Though to integrate higher powers of we use the reduction formula or de moivre theorem, however this stuff isn't in C4.

[zincoff]

(Original post by raheem94)
We usually write it as

I can't think of the answer to your question right now.

The results are usually quoted as 90-x in the books.

That's alright . Thank you!

[ViixKey]

Hey people.
Can someone help with Review Exercise 1, Q8 (Edexcel C3 Book) i know how to do be but i don't know how to use the graph to find the function of f. Thanks

[shahruk]

For solving trigonometric equations which ask for answers in radians, is it ok to do all calculations in degrees and convert the answers to radians at the end ? Thanks !

[Puloot]

help me with this question please!

Use the identities for (sinA + sinB) and (cosA + cosB) to prove that


sin 2x + sin 2y
_______________ = tan (x+y)
cos 2x + cos 2y

[shahruk]

(Original post by ViixKey)
Hey people.
Can someone help with Review Exercise 1, Q8 (Edexcel C3 Book) i know how to do be but i don't know how to use the graph to find the function of f. Thanks

you dont have to find an equation for f(x), find g(root 23) first which is = 5

Now to figure out f(5) you use the graph, find x=5 in the graph and i think you can carry on from here !

[shahruk]

(Original post by Puloot)
help me with this question please!

Use the identities for (sinA + sinB) and (cosA + cosB) to prove that


sin 2x + sin 2y
_______________ = tan (x+y)
cos 2x + cos 2y

SinP + Sin Q = 2* sin[(P+Q)/2 ] * cos[(P-Q)/2 ]

and CosP + Cos Q =2 * cos[(P+Q)/2 ] * cos[(P-Q)/2 ]


P=2x , Q=2y

using this you get


2 sin(x+y) cos(x-y)
--------------------------
2 cos(x+y) cos (x-y)



sin(x+y)
-----------
cos(x+y)


tan(x+y)

[Puloot]

(Original post by shahruk)
SinP + Sin Q = 2* sin[(P+Q)/2 ] * cos[(P-Q)/2 ]

and CosP + Cos Q =2 * cos[(P+Q)/2 ] * cos[(P-Q)/2 ]


P=2x , Q=2y

using this you get


2 sin(x+y) cos(x-y)
--------------------------
2 cos(x+y) cos (x-y)



sin(x+y)
-----------
cos(x+y)


tan(x+y)

Thank youuuu!

[raheem94]

(Original post by zincoff)
That's alright . Thank you!

The range of is

So this method gives,

While the range of is

Hence, we can't do this.

[zincoff]

(Original post by raheem94)


The range of is

So this method gives,

While the range of is

Hence, we can't do this.

Thanks .

[grazie]

(Original post by raheem94)
There is no need to draw a graph.



We know the domain of is

So by looking at we can deduce that the range of is

The range of is the domain of

Hence the domain of is

Hope it makes sense

A useful method that doesn't required a sketch. A little slip at the end though...

The range of is the domain of

Hence the domain of is

[sajidmatin]

Hey people. This things been bugging me.
It says in the C3 book, in the example find


f(x) = f-1 (x)

then it solves it for you, but it just says to make it easier solving it goes straight to f(x) = x

how does this work?

[grazie]

(Original post by sajidmatin)
Hey people. This things been bugging me.
It says in the C3 book, in the example find


f(x) = f-1 (x)

then it solves it for you, but it just says to make it easier solving it goes straight to f(x) = x

how does this work?

Do you mean on p27 where is states...?

When



By definition of the inverse function, f^-1(x) is a reflection of f(x) about the line y=x. Therefore f^-1(x) and f(x) will intersect on the line y=x, so f(x)=x at the point of intersection.

I agree with you that the book should have expanded this to make it more clear. I've attached a graph of the functions used in their example.

Blue :

Red :

Green :

[JenniS]

(Original post by pwcroberts)
None as such but you will need to be able to show all of the double angle formulae and factor formulae from the trig expansions.

Also it is possible you may have to prove some of the standard derrivatives in the formula book by using the derrivatives of sin and cos

okay thanks, think I can do all that (hopefully)!

[hialice]

(Original post by grazie)
Because the range of f(x) is f(x) > 2

It's well worth doing a quick sketch rather than trying to 'plug in' numbers. As stated in in the previous post, I think changing the fraction makes it much more obvious to see where the asymptotes are. So for the inverse function



It's obvious that there's horizontal asymptote at y=1 and a vertical asymptote at x=2.

Thankyou! Makes sense now, I guess I'll just have to remember to long divide in the exam