Asked by Lisa
if x=4 tan theta, express cos(2 theta) as a function of x
Answers
Answered by
Steve
cos 2t = 2cos^2(t) - 1 = 2/sec^2(t) - 1
sec^2 = 1 + tan^2
2/(1 + tan^2(t)) - 1
= [2 - (1 + tan^2 t)]/(1 + tan^2 t)
x = 4tan t
tan t = x/4
[2 - ( 1 + x^2 / 16)]/(1 + x^2 / 16)
(32 - 1 - x^2)/(16 + x^2)
(31 - x^2)/(16 + x^2)
sec^2 = 1 + tan^2
2/(1 + tan^2(t)) - 1
= [2 - (1 + tan^2 t)]/(1 + tan^2 t)
x = 4tan t
tan t = x/4
[2 - ( 1 + x^2 / 16)]/(1 + x^2 / 16)
(32 - 1 - x^2)/(16 + x^2)
(31 - x^2)/(16 + x^2)
Answered by
Steve
The answer above looked wrong to me, and it didn't take long to see that it was so.
cos 2t = 2 cos^2 t - 1
= 2/sec^2 t - 1
= (2 - sec^2 t) / sec^2 t
= (2 - (1 + tan^2 t))/(1 + tan^2 t)
= (1 - tan^2 t)/(1 + tan^2 t)
= (1 - x^2/16)/(1 + x^2/16)
= (16 - x^2)/(16 + x^2)
cos 2t = 2 cos^2 t - 1
= 2/sec^2 t - 1
= (2 - sec^2 t) / sec^2 t
= (2 - (1 + tan^2 t))/(1 + tan^2 t)
= (1 - tan^2 t)/(1 + tan^2 t)
= (1 - x^2/16)/(1 + x^2/16)
= (16 - x^2)/(16 + x^2)
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