Here we use the double angle identity for sine, which is

\(\displaystyle sin(2\alpha) = 2sin\alpha cos\alpha\). 

We can rewrite the originial expression as \(\displaystyle sin\left(\frac{\pi}{6}\right)\) using the double angle identity.

From here we can calculate that

 \(\displaystyle sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\).

One of the double angle formuals for cosine is

\(\displaystyle cos2\alpha = 2cos^2\alpha - 1\). 

We can use this double angle formula for cosine to rewrite the expression given as the \(\displaystyle cos(225^{\circ})\) because \(\displaystyle 2\cdot 112.5 = 225\) and \(\displaystyle \alpha = 112.5\).

We can then calculate that 

\(\displaystyle cos(225^{\circ}) = -\frac{\sqrt2}{2}\).

Here we can use another double angle formula for cosine,

\(\displaystyle cos2\alpha = cos^2\alpha - sin^2\alpha\).

Here \(\displaystyle \alpha = 165^{\circ}\), and so we can use the double angle formula for cosine to rewrite the expression as

\(\displaystyle cos(2\alpha) = cos(2\cdot 165^{\circ}) = cos(330^{\circ})\).

From here we just recognize that 

\(\displaystyle cos(330^{\circ}) = \frac{\sqrt3}{2}\).

Here we can use yet another double angle formula for cosine:

\(\displaystyle cos(2\alpha) = 1 - 2sin^2\alpha\).

First, realize that \(\displaystyle \alpha = 135^{\circ}\).

Next, plug this in to the double angle formula to find that 

\(\displaystyle 1 - 2sin^2(135^{\circ}) = cos(2\cdot 135^{\circ}) = cos(270^{\circ})\).

Here we recognize that \(\displaystyle cos(270^{\circ}) = 0\)

This is a quick test of being able to recall the angle sum formula for sine.

Since, 

\(\displaystyle sin(u\pm v) = sinucov \pm sinvcosu\), and here 

\(\displaystyle u = 120^{\circ} and\: v = 17^{\circ}\), we can rewrite the expression as 

\(\displaystyle sin(120+17) = sin137^{\circ}\).

Which of the following is equivalent to the following expression?

\(\displaystyle \frac{1-sin^2(x)}{csc^2(x)}\)

Recall our Pythagorean trig identity:

\(\displaystyle sin^2(x)+cos^2(x)=1\)

It can be rearranged to look just like our numerator:

\(\displaystyle cos^2(x)=1-sin^2(x)\)

So go ahead and change our original expression to:

\(\displaystyle \frac{cos^2(x)}{csc^2(x)}\)

Then recall the definition of cosecant:

\(\displaystyle csc(x)=\frac{1}{sin(x)}\)

So our original expression can be rewritten as:

\(\displaystyle cos^2(x)\div\frac{1}{sin^2(x)}=cos^2(x)sin^2(x)\)

So our answer is:

\(\displaystyle cos^2(x)sin^2(x)\)

Cosine and sine are not reciprocal functions.  

\(\displaystyle sin (x) = \frac{1}{csc (x)}\) and \(\displaystyle cos (x) = \frac{1}{sec (x)}\)

We begin with the left-hand side of the equation and utilize basic trigonometric identities, beginning with converting the inverse functions to their corresponding base functions:

\(\displaystyle \frac{\frac{1}{\sin^{2}A}}{\frac{1}{\cos^{2}A}} = \cot^{2}A\)

Next, we rewrite the fractional division in order to simplify the equation:

\(\displaystyle \frac{1}{\sin^{2}A}\div\frac{1}{\cos^{2}A} = \cot^{2}A\)

In fractional division we multiply by the reciprocal as follows:

\(\displaystyle \frac{1}{\sin^{2}A}\times \cos^{2}A = \cot^{2}A\)

If we reduce the fraction using basic identities we see that the equivalence is proven:

\(\displaystyle \frac{\cos^{2}A}{\sin^{2}A} = \cot^{2}A\)

\(\displaystyle \cot^{2}A = \cot^{2}A\)

The true identity is \(\displaystyle \small \cos(-x)=\cos(x)\) because cosine is an even function.

The definition of tangent is sine divided by cosine.

\(\displaystyle \tan x=\frac{\sin x}{\cos x}\)