The series expansion of sin θ , cos θ , & tan θ for small values of θ is given by-
sin θ = θ – 
cos θ = 1 – 
tan θ = θ + 
If θ is very very small then the terms containing θ 2, θ 3, .......... and higher powers of θ can be neglected.
∴ For very small θ , sin θ 
θ , cos θ 
1, and tan θ 
θ .
Using the above information answer the following question.
(i) The approximate solution of the equation, sin θ = 0.52 is-
Text Solution
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Ans.
(i)
Sol. sin θ = 0.52 = 0.50 + 0.02 = sin 
, x is verysmall.
⇒ sin θ = sin 
+ cos 
sinx
⇒ sin θ = 
+ 
⇒ 0.52 = 0.5 + 
,
∴ x is very small cosx 
x & sinx 
x
⇒ x = 0.02 × 
radian = 1.32° nearly
⇒ θ = 
= 30° + 1° 19'
= 31°19' approximately.
(ii)
Sol. cos 
= 0.50 – 0.01
⇒ cos 
cos θ – sin 
sin θ = 0.49
⇒ 
= 0.49,
θ is very small so cos θ 
1 sin θ 
θ
⇒ θ = 0.01 × 
radian = 0.66° approximately
⇒ θ = 39' 42" approximately.
(iii)
Sol. 
{by sine rule}
⇒ tanB = 0.05, sinC = sin(90° – B) = cosB
⇒ B 
0.05 radian = 2° 52'36"
∴ C = 90°– 2° 52' 36" = 87° 8' 24".
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