# trig identities q

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

when integrating cos^2 (2x-1), i used the identity cos 2A = 2cos^2A - 1, manipulated it to (1/2)cos (4x - 2) and equalled it - which was the wrong approach. can you not equal them because A has to be like strictly a product of two things? like can it not be i.e. 2x - 1?

0

reply

Report

#2

(Original post by

when integrating cos^2 (2x-1), i used the identity cos 2A = 2cos^2A - 1, manipulated it to (1/2)cos (4x - 2) and equalled it - which was the wrong approach. can you not equal them because A has to be like strictly a product of two things? like can it not be i.e. 2x - 1?

**velaris08**)when integrating cos^2 (2x-1), i used the identity cos 2A = 2cos^2A - 1, manipulated it to (1/2)cos (4x - 2) and equalled it - which was the wrong approach. can you not equal them because A has to be like strictly a product of two things? like can it not be i.e. 2x - 1?

So

and integrate each term when . It helps to see your actual working, you've got the right idea.

Last edited by mqb2766; 1 month ago

0

reply

(Original post by

So

and integrate each term when . It helps to see your actual working, you've got the right idea.

**mqb2766**)So

and integrate each term when . It helps to see your actual working, you've got the right idea.

0

reply

Report

#4

(Original post by

cos 2A / 2 + 1 / 2 ? Where’s the first 2 that’s not 2A coming from? Attaching a picture of my working. It’s not that I don’t get what sol bank says - rather that in the exam I would probably have approached it my way and I don’t get what mistake I’ve made.

**velaris08**)cos 2A / 2 + 1 / 2 ? Where’s the first 2 that’s not 2A coming from? Attaching a picture of my working. It’s not that I don’t get what sol bank says - rather that in the exam I would probably have approached it my way and I don’t get what mistake I’ve made.

Last edited by mqb2766; 1 month ago

0

reply

(Original post by

Your working looks fine and what you've written is the same as in the previous post. Note it helps to put the arguments of the trig function in brackets, so its clear whether operations apply to the argument or to the trig function. If the solution bank says different, can you pls upload it. If you're asking about something else, can you pls clearly state it.

**mqb2766**)Your working looks fine and what you've written is the same as in the previous post. Note it helps to put the arguments of the trig function in brackets, so its clear whether operations apply to the argument or to the trig function. If the solution bank says different, can you pls upload it. If you're asking about something else, can you pls clearly state it.

0

reply

Report

#6

(Original post by

Here’s what sol bank says!

**velaris08**)Here’s what sol bank says!

(cos(2x)-1)^2

not

cos^2(2x-1)

Last edited by mqb2766; 1 month ago

0

reply

(Original post by

Can you upload a picture of the original question. It looks like theyre doing

(cos(2x)-1)^2

not

cos^2(2x-1)

**mqb2766**)Can you upload a picture of the original question. It looks like theyre doing

(cos(2x)-1)^2

not

cos^2(2x-1)

0

reply

Report

#8

(Original post by

“integrate the following” - oh god just realised you’re right, it’s the brackets thing! i mentally added them im so sorry

**velaris08**)“integrate the following” - oh god just realised you’re right, it’s the brackets thing! i mentally added them im so sorry

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top