Trigonometric Identities (1 Viewer)

[skillstriker]

Express sin(theta) and cos(theta) in terms of a, b and c

asin²(theta) + bcos²(theta) = c

 

[Aesytic]

asin^2(theta) + bcos^2(theta) = c
a(1 - cos^2(theta)) + bcos^2(theta) = c
a - acos^2(theta) + bcos^2(theta) = c
[b-a]cos^2(theta) = c - a
cos^2(theta) = (c-a)/(b-a)
.'. cos(theta) = plus minus sqrt{(c-a)/(b-a)}
do the same with with the identity sin^2(theta) + cos^2(theta) = 1 to get sin(theta)

 

[skillstriker]

Thanks!

 

[skillstriker]

Can someone please help me with this question as well:

Prove that (cot*t + cosec*t)² = (1+cost)/(1-cost)

 

[Aesytic]

LHS = (cot(t) + cosec(t))^2
= cot^2(t) + 2cot(t)cosec(t) + cosec^2(t)
using the identity 1 + cot^2(t) = cosec^2(t),
=cot^2(t) + 2cot(t)cosec(t) + 1 + cot^2(t)
= 1 + 2cot^2(t) + 2cot(t)cosec(t)
= 1 + [2cos^2(t)/sin^2(t)] + {2cos(t)/sin(t)}*{1/sin(t)} since cot(t) = cos(t)/sin(t) and cosec(t) = 1/sin(t)
= 1 + [2cos^2(t)/sin^2(t)] + 2cos(t)
= {sin^2(t) + 2cos^2(t) + 2cos(t)}/{sin^2(t)}
= {1 - cos^2(t) + 2cos^2(t) + 2cos(t)}/{1-cos^2(t)}
= {cos^2(t) + 2cos(t) + 1}/{[1-cos(t)]*[1+cos(t)]}
= {[cos(t) + 1]^2}/{[1-cos(t)]*[1+cos(t)]}
= {cos(t)+1}/{1-cos(t)}
= RHS

 

[nightweaver066]

Can someone please help me with this question as well:

Prove that (cot*t + cosec*t)² = (1+cost)/(1-cost)

 

[skillstriker]

Thanks Aesytic and nightweaver for the help.

 

[pokka]

 

[pokka]

oh damn not fast enough XD