What does it mean to use the cosine value of an inverse of tan θ?

1 answer:

Bond [772]3 years ago

3 0

Answer:

$\boxed{ \begin{array}{ccc} \displaystyle \theta = {120}^{ \circ} \\ \displaystyle\sin( {120}^{ \circ} ) = \frac{ \sqrt{3} }{2} \end{array}}$

Step-by-step explanation:

we are given a inverse trig

we want to figure out the value of $\theta$

in that case we can consider using unit circle in unit circle the value of arctan (-√3)=120°

therefore,

the value of theta is 120°

likewise by unit circle the value of sine is

$\displaystyle \sin( {120}^{ \circ} ) = \frac{ \sqrt{3} }{2}$

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zzz [600]

Answer:

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General Formulas and Concepts:

<u>Calculus</u>

Differentiation

Basic Power Rule:

Parametric Differentiation

Integration

Arc Length Formula [Parametric]: $\displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.$

Interval [0, π]

<u>Step 2: Find Arc Length</u>

[Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]: $\displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.$
Substitute in variables [Arc Length Formula - Parametric]: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx$
[Integrand] Simplify: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx$
[Integral] Evaluate: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}$