x Turn on thread page Beta
 You are Here: Home >< Maths

C3 Q's watch

1. Three C3 questions below. Just want to see if i got the same answers.

3a) Binomial

1-x^2 ( all square-rooted)
Up to and including the x^4

b) |lnx|=3

Find the exact values of x.

7a) Identity

cosec^2θ + sec^2θ = cosec^2θsec^2θ

b) 3tan^2θ-5secx +1 =0 between -π<θ(less than or equal)π

Finally

8i) Cartesian

x=3 - 1/sinθ y=2sinθ

All help is appreciated.
2. 3)b)
8)i)

the rest are too long to write up!
3. Anyone try to do question 7a+b?
4. part a) 1 / sin^2x + 1/cos^2x

cos^2x + sin^2x / sin^2xcos^2x

i've started you off there for a, i've re-write cosec and sec in terms of sin and cos. and then cross multiplied

for part b) tan^2x + 1= sec^2x is all you need to know, learn those identities!!!

Spoiler:
Show
a) sin^2x + cos^2x =1
b) 3sec^2x -5secx-2=0 solve this as a quadratic
5. (Original post by devan5)
part a) 1 / sin^2x + 1/cos^2x

cos^2x + sin^2x / sin^2xcos^2x

i've started you off there for a, i've re-write cosec and sec in terms of sin and cos. and then cross multiplied

for part b) tan^2x + 1= sec^2x is all you need to know, learn those identities!!!

Spoiler:
Show
a) sin^2x + cos^2x =1
b) 3sec^2x -5secx-2=0 solve this as a quadratic

See for part a) could you do this as a way:

Take r.h.s

(1 + cos^2x/sin^2x)(1 + sin^2x/cos^2x)

Then times it out to get:

1 + sin^2x/cos^2x + cos^2x/sin^2x + cos^2xsin^2x/cos^2xsin^2x

Which can cancel to give the answer?

(1 + tan^2x) + (1 + cot^2x)
(sec^2x) + (cosec^2x)

??

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 16, 2010
Today on TSR

Negatives of studying at Oxbridge

What are the downsides?

Poll
Useful resources

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

How to use LaTex

Writing equations the easy way

Study habits of A* students

Top tips from students who have already aced their exams

Chat with other maths applicants