
Trigonometry help.
These are not really H.W questions but i have an exam coming very soon and i have some problems:
Trigonometry:
i) Express as a single trigonometric function sinxcosxsec˛x (i keep on going in circles in this one)
ii) A square B, has a diagonal 2x. A square C has twice the area of square A. Find in terms of x, the perimeter of the squre C.
Help would be much appreciated.

i) $\displaystyle sin\,x\, cos\,x\,sec^2x = \frac{sin\,x\,cos\,x}{cos^2\,x} = \left(\frac{sin\,x}{cos\,x}\right) \left( \frac{ {cos\,x}}{ cos\,x} \right)=tan \,x $
ii) A square has 4 equal sides. The diagonal cuts the square into two isosceles, right angled triangles.
Let side of Square B = a. By pythagoras's theorem, $\displaystyle a^2 + a^2 = (2x)^2$
$\displaystyle 2a^2 = 4x^2$
$\displaystyle a^2 = 2x^2$
$\displaystyle a = \sqrt{2}x$
Now Square C has area twice Square B.
Area Square B = $\displaystyle a^2$
Area of Square C = $\displaystyle 2a^2$
From the working above, we know $\displaystyle 2a^2 = 4x^2$
Therefore Area of Square C is (Side C)˛ = $\displaystyle 4x^2$ where Side C is a side of square C
$\displaystyle Side \,C = \sqrt {4x^2} = 2x $
Perimeter = 4(Side C) = 4(2x) = 8x.
Note: Thank you masters for pointing out that I haven't completed the question. Its very late in my time zone, and I misread the question.

If the area of square $\displaystyle C = 4x^2$, then each side = 2x. Therefore, the perimeter would be 4(2x) = 8x

Wow, thanks alot for the quick reply. I totally forgot that
sec˛x=1/cos˛x!