By the double-angula formula for cosine

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

For this problem

\(\displaystyle cos\left[2tan^{-1}\left(\frac{5}{12}\right)\right]=cos^2\left[tan^{-1}\left(\frac{5}{12}\right)\right]-sin^2\left[tan^{-1}\left(\frac{5}{12}\right)\right]\)

\(\displaystyle =\left(\frac{12}{13}\right)^2-\left(\frac{5}{13}\right)^2=\frac{119}{169}\)

By the double-angle formula for the sine function

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

we have

\(\displaystyle x=sin^{-1}\left( \frac{12}{37}\right)\)

thus the double angle formula becomes,

\(\displaystyle sin\left[2sin^{-1}\left(\frac{12}{37}\right)\right]=2sin\left[sin^{-1}\left(12/37\right)\right]cos\left[sin^{-1}\left(12/37\right)\right]\)

\(\displaystyle =2\left(\frac{12}{37}\right)\left(\frac{35}{37}\right)=\frac{840}{1369}\)

The expression \(\displaystyle sin(2\theta)\) is a double angle identity that can also be rewritten as:

\(\displaystyle 2sin(\theta) cos(\theta)\)

Replace the value of theta for \(\displaystyle \theta=X\).

The correct answer is: \(\displaystyle 2sin(X) cos(X)\)

The relevant trigonometric identity is:

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

In this case, "u" is \(\displaystyle 3x\) since \(\displaystyle 2*3x = 6x\).

The only one that actually follows this is \(\displaystyle 1- 2 \sin^2 (3x)\)

A useful trigonometric identity to remember for this problem is 

\(\displaystyle \small 2\cos^2 \theta-1=\cos2\theta\)

or equivalently,

\(\displaystyle \small \small 2\cos^2 \theta=1+\cos2\theta\)

If we substitute \(\displaystyle \small \theta\) for \(\displaystyle \small \frac{\pi}{8}\), we get

\(\displaystyle \small \small \small 2\cos^2 \frac{\pi}{8}=1+\cos2\frac{\pi}{8}=1+\cos\frac{\pi}{4}=1+\frac{\sqrt{2}}{2}\)

A useful trigonometric identity to remember is 

\(\displaystyle \small 2\cos x \sin x = \sin2x\)

If we plug in \(\displaystyle \small \pi/8\) into this equation, we get

\(\displaystyle \small \small 2\cos \left(\frac{\pi}{8}\right)\sin \left(\frac{\pi}{8}\right) = \sin2\left(\frac{\pi}{8}\right)=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}\)

We can divide the equation by 2 to get

\(\displaystyle \small \cos \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2}}{4}\)

Write the half angle identity for sine.

\(\displaystyle sin\left(\frac{A}{2}\right)= \pm\sqrt{\frac{1-cosA}{2}}\)

Since we are given \(\displaystyle sin(30)\), the angle is equal to \(\displaystyle \frac{A}{2}\).  Set these two angles equal to each other and solve for \(\displaystyle A\).

\(\displaystyle \frac{A}{2}=30\)

\(\displaystyle A=60\)

Substitute this value into the formula.

\(\displaystyle sin\left(\frac{60}{2}\right)= \pm\sqrt{\frac{1-cos(60)}{2}}\)

We have:

\(\displaystyle sin^2(a+b)=(sin(a)cos(b)+cos(a)sin(b))^2$$\)

\(\displaystyle =sin^2(a)cos^2(b)+cos^2(a)sin^2(b)+2sin(a)cos(b)cos(a)sin(b)(1)$$\)

 and : 

\(\displaystyle sin^2(a-b)=(sin(a)cos(b)-cos(a)sin(b))^2$$\)

\(\displaystyle =sin^2(a)cos^2(b)+cos^2(a)sin^2(b)-2sin(a)cos(b)cos(a)sin(b)(2)$$\)

(1)-(2) gives:

\(\displaystyle 4sin(a)cos(b)cos(a)sin(b)=[2sin(a)cos(a)] [2sin(b)cos(b)]$$\)

Knowing from the above formula that:( take a=b in the formula above)

\(\displaystyle sin(2a)=2sin(a)cos(a) and sin(2b)=2sin(b)cos(b)$\)

This gives:

\(\displaystyle sin^2(a+b)-sin^2(a-b)=sin(2a)sin(2b)$\)

We note first, using trigonometric identities that: 

\(\displaystyle sin(3a)=sin(a+2a)=sin(a)cos(2a)+cos(a)sin(2a)\)

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

\(\displaystyle sin(2a)=2sin(a)cos(a)\)

This gives:

\(\displaystyle sin(3a)=sin(a)(1-2sin^2(a))+2sin(a)cos^2(a)\)

Since, \(\displaystyle cos^2(a)=1-sin^2(a)\)

We have :

\(\displaystyle sin(3a)=sin(a)-2sin^3(a)+2sin(a)-2sin^3(a)=3sin(a)-4sin^3(a)\)

We can write the above sum as :

\(\displaystyle sin(1)+sin(181)+sin(2)+sin(182)\cdots\)

\(\displaystyle + sin(179)+sin(379) + sin(180)\)

From the given fact, we have :

\(\displaystyle sin(1)+sin(181)=sin(1)+sin(1+180)=sin(1)-sin(1)=0\)

\(\displaystyle sin(2)+sin(182)=sin(2)+sin(2+180)=sin(2)-sin(2)=0\)

\(\displaystyle \vdots\)

\(\displaystyle sin(179)+sin(379)=sin(179)+sin(179+180)=sin(179)-sin(179)=0\)

and we have : \(\displaystyle sin(180)=0\).

This gives :

\(\displaystyle sin(1)+sin(2)+sin(3)\cdots +sin(359)=0\)