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# Urgent P2 help!!! watch

1. I don't know how to do these trig questions:

1) sec^4x - tan^4x = 2sec^2x - 1

2) (cosx + secx)^2 = tan^2x + cos^2x + 3

3) cos^4x - sec^4x = cos^2x - sin^2x

4) cose^2x(tan^2x - sin^2x) = tan^2x

Thanks in advance, the main problem I have with these is that I don't know what to do when there are powers bigger that 2 in the question, so if anyone had any advice I'd be greatful for it.
2. (Original post by Bebop)
I don't know how to do these trig questions:

1) sec^4x - tan^4x = 2sec^2x - 1

2) (cosx + secx)^2 = tan^2x + cos^2x + 3

3) cos^4x - sec^4x = cos^2x - sin^2x

4) cose^2x(tan^2x - sin^2x) = tan^2x

Thanks in advance, the main problem I have with these is that I don't know what to do when there are powers bigger that 2 in the question, so if anyone had any advice I'd be greatful for it.

2) (cosx + secx)^2 = tan^2x + cos^2x + 3

LHS: cos^2x + 2sinxcosx + sin^2x

2sinxcosx = sin2x
cos^2x - sin^2x = cos2x

oh never mind
3. sorry, just try and use the double angle formulaes for them.
4. These question come before the double angle forumla is taught in the textbook so I think you have to use another method.
5. (Original post by Bebop)
3) cos^4x - sec^4x = cos^2x - sin^2x
Is difference of 2 squares so
LHS = (cos^2 x + sin^2 x)(cos^2 x - sin^2 x)
cos^2 x + sin^2 x = 1

The others can be done using cos^2 x + sin^2 x = 1, 1 + tan^2 x = sec^2 x and cot^2 x + 1 = cosec^2 x
6. 2) (cosx + secx)^2 = tan^2x + cos^2x + 3

LHS= cos^2x + 2 + sec^2x
= cos^2x + 2 + (tan^2x + 1)
= tan^2x + cos^2x + 3
= RHS

i'll have the rest up in bout twenty mins haven't got time just now!

sam
x
7. (Original post by Bebop)
I don't know how to do these trig questions:

4) cose^2x(tan^2x - sin^2x) = tan^2x

Thanks in advance, the main problem I have with these is that I don't know what to do when there are powers bigger that 2 in the question, so if anyone had any advice I'd be greatful for it.
I think there's a typo in that one. If you put in x = π/4, you get ¼ = 1!!

cos²x(tan²x - sin²x)
cos²x.sin²x(sec²x - 1)
cos²x.sin²x.tan²x
sin^4x
====
8. (Original post by Fermat)
I think there's a typo in that one. If you put in x = π/4, you get ¼ = 1!!

cos²x(tan²x - sin²x)
cos²x.sin²x(sec²x - 1)
cos²x.sin²x.tan²x
sin^4x
====
I think he means the LHS is: cosec²x(tan²x - sin²x)
anyway, this is how i would do it:
exapnd LHS to get:
cosec²x.tan²x - cosec²x.sin²x

now simplify:

(1/sin²x)(sin²x/cos²x) - (1/sin²x)(sin²x)
(1/cos²x) - 1
sec²x - 1 = tan²x
9. (Original post by Bebop)
I don't know how to do these trig questions:

1) sec^4x - tan^4x = 2sec^2x - 1
LHS: = (sec²x + tan²x)(sec²x - tan²x)
= [sec²x + (sec²x -1)] [sec²x - (sec²x +1)]

= (2sec²x - 1) (1)

= 2sec²x - 1
10. (Original post by mockel)
I think he means the LHS is: cosec²x(tan²x - sin²x)anyway, this is how i would do it:exapnd LHS to get:cosec²x.tan²x - cosec²x.sin²x now simplify: (1/sin²x)(sin²x/cos²x) - (1/sin²x)(sin²x)(1/cos²x) - 1sec²x - 1 = tan²x
How do you get from (1/sin²x)(sin²x/cos²x) - (1/sin²x)(sin²x) to that (1/cos²x) - 1?
11. (Original post by Bebop)
How do you get from (1/sin²x)(sin²x/cos²x) - (1/sin²x)(sin²x) to that (1/cos²x) - 1?
the sin²x cancels for each term
12. (Original post by Bebop)
How do you get from (1/sin²x)(sin²x/cos²x) - (1/sin²x)(sin²x) to that (1/cos²x) - 1?
Mockel meant that the sin^2 x in the denominator cancels out the stn^2 x
in the nemerator in the next term.
13. (Original post by Bebop)
I don't know how to do these trig questions:

1) sec^4x - tan^4x = 2sec^2x - 1

2) (cosx + secx)^2 = tan^2x + cos^2x + 3

3) cos^4x - sec^4x = cos^2x - sin^2x

4) cose^2x(tan^2x - sin^2x) = tan^2x

Thanks in advance, the main problem I have with these is that I don't know what to do when there are powers bigger that 2 in the question, so if anyone had any advice I'd be greatful for it.
ok i can do all of these except for no 3
14. can ne1 do question 3 or is it simply a typo??
15. (Original post by lgs98jonee)
can ne1 do question 3 or is it simply a typo??
See my earlier post (#5)
16. (Original post by Bezza)
Is difference of 2 squares so
LHS = (cos^2 x + sin^2 x)(cos^2 x - sin^2 x)
cos^2 x + sin^2 x = 1

The others can be done using cos^2 x + sin^2 x = 1, 1 + tan^2 x = sec^2 x and cot^2 x + 1 = cosec^2 x
question invloved sec not sin though!!
17. i assume its a typo then
18. (Original post by lgs98jonee)
question invloved sec not sin though!!
Oh yeah. Doesn't work though if it's sec instead of sin. Try x = pi/6 though, cosx = 1/2, secx = 2, sinx = sqrt3/2
cos^4 x - sec^4 x = 1/16 - 16 = -255/16
cos^2 x - sin^2 x = cos(2x) = cos(2pi/3) = -1/2
If you misread it my way then it works!!

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