P2 Trig help please...

Hi, theres a few questions I'm a bit stuck on with some P2 Trig, could anyone help me?

1. f(x) = 3 + 2 sin (2x+k)degrees
where x is between 0 and 360; and k is between 0 and 360.

Show that k = 30, and find the other possible value for k.

2. Given that tan75 = 2 + sqrt3 find in the form m + nsqrt3, where m and n are integers, the value of (i)tan 15 (ii) tan 105

3. Find, in terms of pi, all solutions in interval x is between 0 and 2pi. of

2sin (2x+pi/3) = cos (2x-pi/6)

4. Given that sin(x+@) = sqrt2cos(x-@), where cos x cos @ DOES NOT EQUAL 0

Prove that tanx = (sqrt2-tan@)/(1-sqrt2tan@)

Thanks

Reply 1

evariste_3086

1

StuartYates

Hi, theres a few questions I'm a bit stuck on with some P2 Trig, could anyone help me?

1. f(x) = 3 + 2 sin (2x+k)degrees
where x is between 0 and 360; and k is between 0 and 360.

Show that k = 30, and find the other possible value for k.

2. Given that tan75 = 2 + sqrt3 find in the form m + nsqrt3, where m and n are integers, the value of (i)tan 15 (ii) tan 105

3. Find, in terms of pi, all solutions in interval x is between 0 and 2pi. of

2sin (2x+pi/3) = cos (2x-pi/6)

4. Given that sin(x+@) = sqrt2cos(x-@), where cos x cos @ DOES NOT EQUAL 0

Prove that tanx = (sqrt2-tan@)/(1-sqrt2tan@)

Thanks

not sure what is meant by 1.
2) using tan75 = 2 + sqrt3, tan60=sqrt 3, tan 30= sqrt(3)/3
tan(15)=tan(75-60)=tan75-tan60/(1+tan75tan60)=
2/(4+2sqrt(3)) now mult top and bottom by 4-2sqrt(3) to get
tan 15=2-sqrt(3)

tan 105=tan (75+30)=tan 75+tan 30/(1-tan75tan30)
={2+4/3sqrt(3)}/-2/3sqrt(3)
mult top and bottom by sqrt(3) to get
tan(105)=-2-sqrt(3)

2sin (2x+pi/3) = cos (2x-pi/6)
2{sin2xcospi/3+cos2xsinpi/3}=cos2xcospi/6+sin2xsinpi/6
sin2x+sqrt(3)cos2x=1/2{sqrt(3)cos2x+sin2x}
2sin2x+2sqrt(3)cos2x=sqrt(3)cos2x+sin2x
sin2x=-sqrt(3)cos2x
tan2x=-sqrt(3)
2x=120,300,480,660
x=60,150,240,330
I would check these are all. I do tend to miss some!!
Given that sin(x+a) = sqrt2cos(x-a), where cos x cos @ DOES NOT EQUAL 0
sin(x+a)=
sinxcosa+cosxsina=sqrt2{cosxcosa+sinxsina}
so tanxcosa+sina=sqrt(2){cosa+tanxsina}....divide by cosx
tanx+tana=sqrt(2){1+tanxtana}.......divide by cos a
tanx(1-sqrt(2)tana)=sqrt(2)-tana
hence tanx=(sqrt2-tana)/(1-sqrt2tana)

Reply 2

StuartYates

OP

2

Hi, I've started this question, and am stuck part way through. could someone give me tips for finishing it off. (Cheers for the above reply by the way, most helpful!)

Find, to the nearest integer, the values of x in the interval x= something between 0 and 180 for which

3sin^2x-7cos3x-5=0

so far i've got:

3(1-cos^2 3x) - 7cosx -5 =0
3-(3cos^2 9x) - 7cosx -5= 0
(3cos^2 9x) + 7cosx + 2 = 0

Any help anyone is appreciated, thanks!

Reply 3

El Stevo

11

3sin^2x-7cos3x-5=0

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

sub in 1-(cos^2)x for (sin^2)x

3(1-cos^2)x-7cos3x-5=0

3 - 3(cos^2)x - 7cos(3x) - 5 = 0

-3(cos^2)x - 7cos(3x) - 2 = 0

3(cos^2)x + 7cos(3x) + 2 = 0

use double angle formula twice to work out cos(3x) in terms of x... or De Moivres...

Reply 4

StuartYates

OP

2

Hi again, thanks for the help above. A little bit stuck on another few.

1. tan x = (sqrt2-tan@)/(1-sqrt2tan@). Hence, or otherwise, find solutions in the interval 0 to 2pi of:

sin(x+(pi/6))=sqrt2cos(x-(pi/6))

giving your answer in radians.

2. Solve, giving your answers in terms of pi:

cos2x+3sinx=2, for the range x= 0 to 2pi

thanks for your help everyone!!!

OP

anyone please?

for number two bung in the double angle formula (for cos(2x))

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

(cos^2)x - (sin^2)x + 3sinx = 2

using (cos^2)x + (sin^2)x = 1

1 - (sin^2)x - (sin^2)x + 3sinx = 2

2(sin^2)x - 3sinx +1 = 0

let (sin^2)x = y

2(y^2) - 3y + 1 = 0

(2y-1)(y-1) = 0

y= (1/2)
y = 1

cosx = (1/2)
cosx = 1

you should be able to do it from here... :smile:

OP

Fantastic, thanks!

StuartYates

Hi again, thanks for the help above. A little bit stuck on another few.

1. tan x = (sqrt2-tan@)/(1-sqrt2tan@). Hence, or otherwise, find solutions in the interval 0 to 2pi of:

sin(x+(pi/6))=sqrt2cos(x-(pi/6))

giving your answer in radians.

2. Solve, giving your answers in terms of pi:

cos2x+3sinx=2, for the range x= 0 to 2pi

thanks for your help everyone!!!

1) sin(x+(pi/6))=sqrt2cos(x-(pi/6)) so
tan x=(sqrt2-tan(pi/6))/(1-sqrt2tan(pi/6)). by first post
x= 77.63, 257.63 i leave for you to put in radians

2)cos2x+3sinx=2,
COS2X=COS^2X-SIN^2X=(1-SIN^2X)-SIN^2X=1-2SIN^2X]
so 1-2sin^2x+3sinx=2
ie 2s^2-3s+1=o (s=sinx)
(2s-1)(s-1)=0
so sinx =1
sinx =1/2
etc

Reply 9

evariste_3086

1

El Stevo