Hi, if Cos 12° =h, find Sin12°in terms of h...subsequently, write down the value of Cot 12°...please help
+343 \(cos^2x+sin^2x=1<=> h^2 + sin^212=1<=>sin12= \sqrt{1-h^2}=\sqrt{(1-h)(1+h)}\)
\(cot12=\frac{cos12}{sin12}=\frac{h}{\sqrt{(1-h)(1+h)}}\)
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+9479 By the Pythagorean Identity,
\(\sin^2(12^\circ)+\cos^2(12^\circ)\ =\ 1\)
We are given that cos(12°) = h so we can substitute h in for cos(12°)
\(\sin^2(12^\circ)+h^2\ =\ 1\)
Subtract h2 from both sides of the equation.
\(\sin^2(12^\circ)\ =\ 1-h^2\)
Because 12° is in Quadrant I, sin(12°) is positive. So take positive sqrt of both sides.
\(\sin(12^\circ)\ =\ \sqrt{1-h^2}\)
By definition of cotangent,
\(\cot(12^\circ)\ =\ \frac{\cos(12^\circ)}{\sin(12^\circ)}\)
Substitute h in for cos(12°) and substitute √[ 1 - h2 ] in for sin(12°)
\(\cot(12^\circ)\ =\ \frac{h}{\sqrt{1-h^2}}\)_
