trig identity question (1 Viewer)

[izi]

a) Prove: tan x - tan y / tan x + tan y = sin (x-y) / sin (x+y)


b) If 4 cos A + 3 = 0 and tan A > 0, find the exact value of:
(i) tan (A - 180 degrees)
(ii) cos (A + 90 degrees)

c) (i) Hence or otherwise prove that cos3Q = 4cos^3Q-3cosQ (Q is theata)
(ii) Solve 8cos^3Q - 6cosQ - /\3 = 0 for 0degrees < Q < 360degrees (/\ is square root)

thanx

 

[Li0n]

a) Prove: tan x - tan y / tan x + tan y = sin (x-y) / sin (x+y)

RHS = (sinxcosy - cosxsiny)/sinxcosy + cosxsiny
LHS = (sinx/cosx - siny/cosy)/(sinx/cosx + siny/cosy)

= [(sinxcosy - cosxsiny)/cosxcosy] / [(sinxcosy + sinycosx)/cosxcosy]
cancel out the cosxcosy
:. = (sinxcosy - cosxsiny)/sinxcosy + cosxsiny
= RHS

ahh for the other two, i cbf + i should study english now

 

[velox]

isnt izi in Uni?

 

[izi]

yes, izi is at uni. im her brother (class of 2005).

 

[velox]

ah ok...same as me then

 

[Funky_Junky]

well i wrote out the full working out for question 3 coz typing it down got too complicated

left the last bit for you to do coz i can't be bothered to go and get a calculator...if you are still stuck lemme know and i will expand it for you.

-funky

 

[Funky_Junky]

(ii)
CosA= -3/4
TanA=root 7/3

tan (A-180 degrees)
= tanA-tan180/1+tanAtan180

**tan 180=0**

tanA/1

TanA=root7/3

 

[Xayma]

b) If 4 cos A + 3 = 0 and tan A > 0, find the exact value of:
(i) tan (A - 180 degrees)
(ii) cos (A + 90 degrees)

c) (i) Hence or otherwise prove that cos3Q = 4cos^3Q-3cosQ (Q is theata)
(ii) Solve 8cos^3Q - 6cosQ - /\3 = 0 for 0degrees < Q < 360degrees (/\ is square root)

thanx

b) i)cos A=-3/4
&there4; tan A=-&radic;7/-3
tan A=&radic; 7/3

&there4; tan (A-180&deg; )=tan A
=&radic; 7/3

ii) cos [A+90&deg;]=cos A*cos 90&deg;-sin A*sin 90&deg;
=-sin A
Now sin A=-&radic;7/4
&there4; cos[A+90&deg;]=&radic;7/4

c) i) cos 3&theta;=cos (2&theta;+&theta;.)
=cos 2&theta;cos&theta;-sin2&theta;*sin&theta;
=2cos³&theta;-cos&theta;-2sin²&theta;cos&theta;
=2cos³&theta;-cos&theta;-2(1-cos²&theta;.)cos&theta;
=4cos³&theta;-3cos&theta;

ii) You shouldn't need a calc from Funky Junky cos3&theta;=&radic;3/2
3&theta;=&pi;/6, 5&pi;/6, 7&pi;/6, 11&pi;/6, 13&pi;/6, 17&pi;/6
&theta;=&pi;/18, 5&pi;/18, 7&pi;/18, 11&pi;/18, 13&pi;/18, 17&pi;/18

 

[Funky_Junky]

sinA= -root 7/4

cos (A+90)
cosAcos90-sinAsin90
-sinA

-sinA= - root7/4
sinA=root7/43

 

[Funky_Junky]

Well i have not learnt radians at skool yet...