Answer:Let =tan2
t
=
tan
x
2
, then cos=1−21+2
cos
x
=
1
−
t
2
1
+
t
2
and tan=21−2
tan
x
=
2
t
1
−
t
2
.
Now
1+21−2+21−2(1+)21−21+1−tan2
Step-by-step explanation:
true if that's what ur looking for. hope this helps
Answer:
The two iterations of f(x) = 1.5598
Step-by-step explanation:
If we apply Newton's iterations method, we get a new guess of a zero of a function, f(x), xₙ₊₁, using a previous guess of, xₙ.
xₙ₊₁ = xₙ - f(xₙ) / f'(xₙ)
Given;
f(xₙ) = cos x, then f'(xₙ) = - sin x
cos x / - sin x = -cot x
substitute in "-cot x" into the equation
xₙ₊₁ = xₙ - (- cot x)
xₙ₊₁ = xₙ + cot x
x₁ = 0.7
first iteration
x₂ = 0.7 + cot (0.7)
x₂ = 0.7 + 1.18724
x₂ = 1.88724
second iteration
x₃ = 1.88724 + cot (1.88724)
x₃ = 1.88724 - 0.32744
x₃ = 1.5598
To four decimal places = 1.5598