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Simplify

Simplify: 5sin23theta +5cos23theta and (1+sinx)2 + (1-sinx)2 + 2cos2x can you please show me step by step... :tongue:
(edited 11 years ago)

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Reply 1
Original post by SophieL1996
Simplify: 5sin23theta +5cos23theta (1+sinx)2 + (1-sinx)2 + 2cos2x can you please show me step by step... :tongue:


What have you done?
Reply 2
Original post by BabyMaths
What have you done?


I cant do these ones
Reply 3
Original post by SophieL1996
I cant do these ones


You can do something with it surely.

It doesn't matter if what you try first isn't the best idea. You just have to try!

Edit: neg. lol.

My advice was well meant and it is good advice.
(edited 11 years ago)
Reply 4
Original post by BabyMaths
You can do something with it surely.

It doesn't matter if what you try first isn't the best idea. You just have to try!


Yes I have tried... but I cannot get the right answer, I just need someone to do the steps for me so I can have a go at some other ones...
Reply 5
Original post by SophieL1996
Simplify: 5sin23theta +5cos23theta (1+sinx)2 + (1-sinx)2 + 2cos2x can you please show me step by step... :tongue:



Original post by SophieL1996
I cant do these ones


Have you even simplified the part I have highlighted
Reply 6
Original post by SophieL1996
I just need someone to do the steps for me so I can have a go at some other ones...


If this were the case then you would be able to use the examples in your text or those that your teacher has given you
Reply 7
Original post by SophieL1996
Yes I have tried... but I cannot get the right answer, I just need someone to do the steps for me so I can have a go at some other ones...


Expand the two brackets, that should cancel some terms, then you need to use the rule sin2θ+cos2θ=1sin^2\theta + cos^2\theta=1 (same can apply for x terms btw, not just theta)
(edited 11 years ago)
Reply 8
Original post by TenOfThem
If this were the case then you would be able to use the examples in your text or those that your teacher has given you


My teacher didb't really go into any of it and the examples are a lot easier than these questions, so I am stuck. Any help would be appreciated. :smile:
Reply 9
Original post by SophieL1996
Yes I have tried... but I cannot get the right answer, I just need someone to do the steps for me so I can have a go at some other ones...


I have just realised these are 2 different questions

The one that I highlighted is very straightforward ... what have you tried

Then 5sin2A=5cos2A=5(sin2A+cos2A)5\sin ^2A = 5\cos ^2A = 5(\sin ^2A + \cos ^2A)

And you should certainly know what sin2A+cos2A=\sin ^2A + \cos ^2A =
Reply 10
Original post by Robbie242
Expand the two brackets, that should cancel some terms, then you need to use the rule sin2θ+cos2θ=1sin^2\theta + cos^2\theta=1
So far I have got 2 + 2sinx2+2cos2x
where do I go from there?
Original post by SophieL1996
My teacher didb't really go into any of it and the examples are a lot easier than these questions, so I am stuck. Any help would be appreciated. :smile:


Since these only require you to

Multiply Brackets
Know that sin2A+cos2A=1\sin ^2A + \cos ^2A = 1
Original post by SophieL1996
So far I have got 2 + 2sinx2+2cos2x
where do I go from there?


Are you saying that you do not know what sin2A+cos2A=\sin ^2A + \cos ^2A =
Reply 13
Original post by SophieL1996
So far I have got 2 + 2sinx2+2cos2x
where do I go from there?


Take out a factor of 2 and follow a similar process to what TenOfThem has done hint: sin2θ+cos2θ=1sin^2\theta + cos^2\theta=1
(edited 11 years ago)
Reply 14
Original post by TenOfThem
I have just realised these are 2 different questions

The one that I highlighted is very straightforward ... what have you tried

Then 5sin2A=5cos2A=5(sin2A+cos2A)5\sin ^2A = 5\cos ^2A = 5(\sin ^2A + \cos ^2A)

And you should certainly know what sin2A+cos2A=\sin ^2A + \cos ^2A =


So far I have got 2 + 2sinx2+2cos2x
Reply 15
Original post by SophieL1996
Simplify: 5sin23theta +5cos23theta and (1+sinx)2 + (1-sinx)2 + 2cos2x can you please show me step by step... :tongue:


Why didn't you the and in in the first place?
Original post by SophieL1996
So far I have got 2 + 2sinx2+2cos2x


And, as I have said

Do you know what sin2x+cos2x=\sin ^2x + \cos ^2x =
Reply 17
Original post by TenOfThem
Are you saying that you do not know what sin2A+cos2A=\sin ^2A + \cos ^2A =


it equals 1...
Reply 18
Original post by SophieL1996
it equals 1...
So what would 2 times it equal to?
Reply 19
Original post by Robbie242
So what would 2 times it equal to?


2:smile:

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