1. ## Trigonometry.

Alevel (module 2/3 standard). Any help at all would be hugely appriciated.

sin (x + pi/6) = 2cosx

giving all solutions in range of 0 < x < 2pi

b) Use the formular tan (a - b) = (tana - tanb)/(1 + tana tanb)

to show that tan(x - pi/4) = (tanx - 1)/(tan x+1)

c) Hence or otherwise solve in radians tan(x - pi/4) = 6tanx

67. Betweem -180 < x < 180 solve

cosec^2 x = 1 + 2 sec^2 x

68. For all values of x prove that

8 sin^4 x = 3 - 4 cos 2 x + cos 4 x

Hence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0

70. Find the values of x for 0-360 for which tan^2 x + 2 sec x = 5

85. Find values to which

3 cos 2x + 7 cos x = 0 and 0 < x < 2pi

2. Originally Posted by Sarah.
Alevel (module 2/3 standard). Any help at all would be hugely appriciated.

sin (x + pi/6) = 2cosx

giving all solutions in range of 0 < x < 2pi
Hint: use the identity $\displaystyle \sin (A + B) = \sin A \cos B + \sin B \cos A$ and see where you get

3. Originally Posted by Sarah.

b) Use the formular tan (a - b) = (tana - tanb)/(1 + tana tanb)

to show that tan(x - pi/4) = (tanx - 1)/(tan x+1)
just use the formula:

$\displaystyle \tan \left( x - \frac {\pi}4 \right) = \frac {\tan x - \tan \frac {\pi}4 }{1 + \tan x \tan \frac {\pi}4 }$ .........i trust you know what $\displaystyle \tan \frac {\pi}4$ is

c) Hence or otherwise solve in radians tan(x - pi/4) = 6tanx
from above, we know that $\displaystyle \frac {\tan x - 1}{\tan x + 1} = 6 \tan x$

begin by solving for tanx, then take the arctan (that is, the inverse tangent) of both sides to find the required x-values

4. Originally Posted by Sarah.

67. Betweem -180 < x < 180 solve

cosec^2 x = 1 + 2 sec^2 x
use the fact that:

$\displaystyle \csc^2 x = 1 + 2 \sec^2 x$

$\displaystyle \Rightarrow \frac 1{\sin^2 x} = 1 + \frac 2{\cos^2 x}$

combine the fractions on the right and see if you can take it from there

5. Originally Posted by Sarah.
68. For all values of x prove that

8 sin^4 x = 3 - 4 cos 2 x + cos 4 x

usually i'd tell you to start working on the right, but maybe it's better to start on the left here

note,
$\displaystyle 8 \sin^4 x = 8 \left( \sin^2 x \right)^2 = 8 \left( 1 - \cos^2 x \right)^2$

now expand and simplify to get the right hand side

(remember your formulas for the double angle of cosine)
Hence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0
in light of the above, we have $\displaystyle 4 - 8 \sin^4 x = 0$

can you continue?

6. 32 was fine thanks.

67. [tex]
1/(sin^2x) = 1 + 2/(cos^2x)

After this, I got a bit stuck. Another hint would be useful please.

68.
Quote:
Hence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0
in light of the above, we have

can you continue?

In short, no. I'm not 100% sure where you got the first line from, and from there how to solve the final equation.

Thanks for the help so far.

7. Originally Posted by Sarah.
32 was fine thanks.

67. [tex]
1/(sin^2x) = 1 + 2/(cos^2x)

After this, I got a bit stuck. Another hint would be useful please.
that's where i left off. you did nothing after that? do you know how to combine fractions?

you would have:
$\displaystyle \frac 1{\sin^2 x} = \frac {\cos^2 x + 2}{\cos^2 x}$

do you think you can continue now?

68.
Quote:
Hence find in radians in terms of pi the values of x in pi and -pi for which cos 4x - 4cos 2x + 1 = 0
in light of the above, we have

can you continue?

In short, no. I'm not 100% sure where you got the first line from, and from there how to solve the final equation.

Thanks for the help so far.
remember, at this point we showed that $\displaystyle 8 \sin^4 x = 3 - 4 \cos 2x + \cos 4x$

$\displaystyle \Rightarrow -4 \cos 2x + \cos 4x = 8 \sin^4 x - 3$ (i had a typo before, i thought it was - cos (4x) and + 4cos(2x))

add 1 to both sides, we get:

$\displaystyle \cos 4x - 4 \cos 2x + 1 = 8 \sin^4 x - 2$

now we are told that $\displaystyle \cos 4x - 4 \cos 2x + 1 = 0$

thus we have:

$\displaystyle 0 = 8 \sin^4 x - 2$

now continue

8. Originally Posted by Sarah.

70. Find the values of x for 0-360 for which tan^2 x + 2 sec x = 5

remember, $\displaystyle \tan^2 x = \sec^2 x - 1$

thus,

$\displaystyle \tan^2 x + 2 \sec x = 5$

$\displaystyle \Rightarrow \sec^2 x - 1 + 2 \sec x = 5$

bring everything to one side, we have:

$\displaystyle \sec^2 x + 2 \sec x - 6 = 0$

this is quadratic in $\displaystyle \sec x$, if we were to replace $\displaystyle \sec x$ with $\displaystyle y$, for instance, we would have

$\displaystyle y^2 + 2y - 6 = 0$

i think you can solve that for $\displaystyle y$. when done, replace $\displaystyle y$ with $\displaystyle \sec x$ and solve for $\displaystyle x$

9. Originally Posted by Sarah.
85. Find values to which

3 cos 2x + 7 cos x = 0 and 0 < x < 2pi
recall that $\displaystyle \cos 2x = 2 \cos^2 x - 1$, thus we have:

$\displaystyle 6 \cos^2 x - 3 + 7 \cos x = 0$

$\displaystyle \Rightarrow 6 \cos^2 x + 7 \cos x - 3 = 0$

and it is a problem like #70 from there

10. 67. So you end up with:

cos^2x = (sin^2x)(cos^2x + 2)

So you divide by cos^2 x?

And get

tan^2 x + 2 tan^2 x -1

3 tan^2 x -1?

Or have I done something wrong?

What do you do here next?

68. How do you simplify/factorise this?

11. Originally Posted by Sarah.
67. So you end up with:

cos^2x = (sin^2x)(cos^2x + 2)

So you divide by cos^2 x?

And get

tan^2 x + 2 tan^2 x -1

3 tan^2 x -1?

Or have I done something wrong?

What do you do here next?
i do not recommend dividing by cos^2 x. if a solution of x ends up being one that causes cos(x) = 0, it would mean you have divided by 0, hence making your solution invalid. bring everything to one side and factor out the cos^2 x instead

68. How do you simplify/factorise this?
no factorization necessary, just solve for sin^4 x and take it from there

12. Doesn't it end up being
= sin^2 x cos^2 x + 2sin^2x - cos^2x

So cos^2x isn't common through out the equation and so cannot take out as a factor?

13. Oh wait. for 67, the original problem has sec^2 x. that means cos^2 x in not zero, since sec^2 x would be undefined then. thus it was ok to divide by cos^2 x

14. Originally Posted by Sarah.
Doesn't it end up being
= sin^2 x cos^2 x + 2sin^2x - cos^2x

So cos^2x isn't common through out the equation and so cannot take out as a factor?
you could have taken out cos^2 x out of the terms where it's common. but i made a post saying why it is ok to divide by cos^2 x, so don't worry about it

EDIT: i'm kind of busy with other stuff right now, and as a result, i cannot concentrate on your problems as much as i would like to. i'm going to log off. hopefully someone else will help you. if not, i'll pick up where i left off when i come back on again

15. Thank you very much for sticking through this with me, I know how tedious it must be. Enjoy the rest of your night.

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