Trigonometry.

**Sarah.** · December 2nd 2007, 11:42 AM

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

32. a) Solve in radians

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

Thank you for reading. xx

**Jhevon** · December 2nd 2007, 11:44 AM

Originally Posted by Sarah.

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

32. a) Solve in radians

sin (x + pi/6) = 2cosx

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

Hint: use the identity $\sin (A + B) = \sin A \cos B + \sin B \cos A$ and see where you get

**Jhevon** · December 2nd 2007, 11:49 AM

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:

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

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

from above, we know that $\frac {\tan x - 1}{\tan x + 1} = 6 \tan x$

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

**Jhevon** · December 2nd 2007, 11:53 AM

Originally Posted by Sarah.

67. Betweem -180 < x < 180 solve

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

use the fact that:

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

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

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

**Jhevon** · December 2nd 2007, 12:09 PM

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

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

can you continue?

**Sarah.** · December 2nd 2007, 01:00 PM

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.

**Jhevon** · December 2nd 2007, 01:40 PM

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

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

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

add 1 to both sides, we get:

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

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

thus we have:

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

now continue

**Jhevon** · December 2nd 2007, 01:46 PM

Originally Posted by Sarah.

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

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

thus,

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

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

bring everything to one side, we have:

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

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

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

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

**Jhevon** · December 2nd 2007, 01:49 PM

Originally Posted by Sarah.

85. Find values to which

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

recall that $\cos 2x = 2 \cos^2 x - 1$ , thus we have:

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

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

and it is a problem like #70 from there

**Sarah.** · December 2nd 2007, 02:04 PM

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?

**Jhevon** · December 2nd 2007, 02:17 PM

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

**Sarah.** · December 2nd 2007, 02:31 PM

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?

**Jhevon** · December 2nd 2007, 02:31 PM

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

**Jhevon** · December 2nd 2007, 02:32 PM

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

**Sarah.** · December 2nd 2007, 02:34 PM

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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