Pythagorean Identities

Simplify \(\sin x \cos^2 x = \sin x -1\) and find the value of \(x\): \(0 ＜ x ＜ 2\pi\).

For this, we will need to use the first Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1\) and rearrange it:

\[ \cos^2 x = 1 - \sin^2 x .\]

We can now substitute \(1 - \sin^2 x \) into the expression:

\[ \sin x \cos^2 x = \sin x(1 - \sin^2 x ).\]

Simplifying this and setting it equal to the right hand side, we get

\[ \sin x - \sin^3 x = \sin x -1 \]

or

\[-\sin^3 x = -1. \]

So \( \sin x = 1 \) and \(x = \frac{\pi}{2}\).

If \(\cos x = 0.78\), what is the value of \(\tan x\)?

For this, we need to use the fact that \( \tan^2x + 1 = \sec^2x \). We also know that

\[ \sec x = \frac{1}{\cos x}\]

therefore

\[ \sec x = \frac{1}{0.78} = 1.282 .\]

We can now substitute this value into the equation and find \( \tan x\):

\[ \tan^2 x + 1 = (1.282)^2 \]

so

\[ \tan^2 x = (1.282)^2 -1 \]

and \( \tan x = 0.802\).

Solve for \(x\) between \(0^\circ\) and \(180^\circ\):

\[ \cot^2 (2x)+ \csc (2x) - 1 = 0.\]

In this case, we need to use the third Pythagorean identity, \( 1 + \cot^2\theta = \csc^2\theta\).

If we rearrange this identity, we get \( \cot^2\theta = \csc^2\theta - 1\). In this case \(\theta = 2x\) and we can plug in this rearranged identity into our equation:

\[ \left( \csc^2(2x) - 1 \right) + \csc 2x - 1 = 0 \]

so

\[ \csc^2 2x + \csc 2x - 2 = 0.\]

We can treat this as a quadratic that we can factorise into

\[(\csc 2x + 2)(\csc 2x - 1) = 0.\]

We can now solve this and get \( \csc 2x = -2\) or \( \csc 2x = 1\), so \( \sin 2x = -\frac{1}{2}\) or \(\sin x = 1\). Therefore \(2x = 210^\circ\), \(330^\circ\), \(90^\circ\). and \(x = 45^\circ\), \(105^\circ\), \(165^\circ\).