maths c2 edexcel

[astarmathsandphysics]

are they cos^2 x sin^2 x etc

[halosh92]

solve the following equations for thita, in the interval 0<thita<360

a) sin thita + cos thita = 0

b) (sin thita - 1)(5 cos thita + 3) = 0

solve the following equations for "x" , giving your answers to 3sf where appropriate, in the intervals indicated:

a) tan x = - sqrt3 /3             0<x<720
b) tan x = 2.90                    80<x<440


solve in the intervals indicated . the following equations for thita, where thita is measured in radians,
give your answer in terms of pie or 2pie or 2 decimal places.

a)  sin thita =0         -2pie<thita<2pie

b) 2 cos thita = 3 sin thita       0<thita<2pie

[halosh92]

yep
sorry typing error

[astarmathsandphysics]

a)sin x +cos x=0
sin x =-cos x
tan x =-1
x=-45,180-45, 360-45 etc

[astarmathsandphysics]

(sin x-)(5cos x+3)=0
sinx-1=0 so sinx =1 so x=90, 180-90, 360+90, etc
only 90 in range
5cos x+3=0
cos x=-3/5 so x=126.9
x=126.9,360-126.9, 360+126.9, etc
only 1st two in range

[astarmathsandphysics]

for the a and b tan questions find any answer to tanx=c then keep adding 180

sinx=0 x=-360,-180,0,180,360

2cos x=3 sinx so tanx=2/3 so x=33.7,180+33.7,360+33.7 etc

Can you post the very first question with all the squares made clear

[halosh92]

for the a and b tan questions find any answer to tanx=c then keep adding 180

sinx=0 x=-360,-180,0,180,360

2cos x=3 sinx so tanx=2/3 so x=33.7,180+33.7,360+33.7 etc

Can you post the very first question with all the squares made clear

for any tan questions !!-do we ALWAYS add 180???

[astarmathsandphysics]

yes always add 180
http://www.astarmathsandphysics.com/a_level_maths_notes/C3/a_level_maths_notes_c3_finding_multiple_solutions_of_trigonometric_equations.html

[halosh92]

okay thx alot 

binomial expansion :

(when (1-3/2 x)^p is expanded in ascending powers of "x" , the coefficient of "x" is -24
a) find the value of "p"
b) find the coefficient of x^2 in the expansion
c) find the coefficient of x^3 in the expansion

[astarmathsandphysics]

(1-3x/2)^p =1+(-3px/2)+(p(p-1)/2!*(-3x/2)^2+(p(p-1)(p-2)/3!)(-3x/2)^3+....
-3p/2=-24 so p=16
coefficienct of x^2 (p(p-1)/2!*(-3/2)^2 =(16(16-1)/2!*(-3/2)^2 =270
coefficient of x^3 (p(p-1)(p-2)/3!)(-3/2)^3 =(16(16-1)(16-2)/3!)(-3/2)^3 =1260

[halosh92]

the vertices of a quadrilateral ABCD has coordinates A(-1,5)     B(7,1)       C(5,-3)           D(-3,1)
show that the quadrileteral is a triangle

[astarmathsandphysics]

You mean a parallelogram. In morning

[halosh92]

You mean a parallelogram. In morning

sorry i meant rectangle not paralleolgram
 

[astarmathsandphysics]

A(-1,5)     B(7,1)       C(5,-3)           D(-3,1)
AB=(7,1)-(-1,5)=(8,-4)
C-D=(5,-3)-(-3,1)=(8,-4)=B-A so Ab is same length as CD
CB=(7,1)-(5,-3)=(2,4)=DA
Also AB.CB=8*2+-4*4=0 etc

[halosh92]

find the coordinates of the point where the tangent to the curve y=x^2 + 1  at the point (2,5)
meets the normal to the same curve at the point (1,2)