Trigonometry (1 Viewer)

[M@ster P]

Use the t results to:

i) prove that if tan@/2 = b/a, then acos@ +bsin@ = a

ii) Show that if sec@ - tan@ = x, then x = (1-t)/(1+t) where t = tan@/2. Thus prove t = (1-x)/(1+x)

iii) If t = tanA/2, prove that cosA = (1- t^2)/(1+ t^2). Also if tanB= 0.5t, and cosC = (5cosA + 3)/(3cosA + 5) where A, B, C are acute angles, prove that angle A = twice angle B.


Thanks

 

[azureus88]

(i) LHS = acos@ + bsin@
= a[(1-t^2)/(1+t^2)] + b[2t/(1+t^2)] where t = tan(@/2) = b/a
= a[1-(b/a)^2)/(1+(b/a)^2)] + b[2(b/a)/(1+(b/a)^2)]
= a[(a^2 - b^2)/(a^2 + b^2)] + b[2ab/(a^2 + b^2)
= [a(a^2 + b^2)] / (a^2 + b^2)
= a
= RHS

(ii) LHS = [(1+t^2)/(1-t^2)] - [2t/(1-t^2)]
= [(1-t)^2]/(1-t^2)
= [(1-t)^2])]/[(1-t)(1+t)]
= (1-t)/(1+t)
= x

(1-t)/(1+t) = x
x(1+t) = (1-t)
xt+t = 1-x
t(1+x) = 1-x
t = (1-x)/(1+x)

(iii) first part, just draw a right angle triangle with angle tanA/2 and sides t, 1, and sqrt(1+t^2), then work out cosA/2 and sinA/2 and use that to work out cosA using double angle forumla.

dont no how to do last part. sry

 

[M@ster P]

thx

for part two where you proved that t = (1-x)/(1+x). I don't really understand how you manipulated it to make t the subject, can someone mind explaining it to me

 

[azureus88]

ok, we proved part 1 of question 2 that x=(1-t)/(1+t)

now, multiply both sides by (1+t), we get... x(1+t)=(1-t)
expanding LHS... x+xt=1-t
rearranging... xt+t=1-x
factorising... t(1+x)=1-x
then dividing by (1+x) u get ur answer

 

[M@ster P]

lol thanks man, it was so simple, I think one your previos post you missed out a line a working, made me confused.