Thread: Dreaded identities and double angle formulae

1. Dreaded identities and double angle formulae

I've always had problems with these. Surely it cant be that hard, but I really cant get my head around them as it stands, even if it is just proving them - Maybe because I cant find many practical uses for it.. Anyway, these are the following coursework questions due for tomorrow. Would be very greatful if you could provide full answers so I can actually start to understand these damned things. Cheers!

[Y=Theta]

12) Using double angle formulae rewrite $\displaystyle sin 4Y$ in terms of $\displaystyle sin Y$ and $\displaystyle cos Y$ (Raised to any power).

13) Prove the following identitys:

a) $\displaystyle cot^2 x+1 = cosec^2 x$

b) $\displaystyle (tan x+sec x)/(sec x-cos x+tan x) = cosec x$

c) $\displaystyle sin^3 x-cos^3 x = (1+sin x*cos x)(sin x-cos x)$

For c) we were told to go about doing the right hand side first.

Thanks again!

2. Trig Identities

Hello Backlock
Originally Posted by Backlock
[Y=Theta]

12) Using double angle formulae rewrite $\displaystyle sin 4Y$ in terms of $\displaystyle sin Y and cos Y$ (Raised to any power).
$\displaystyle \sin 4\theta = \sin (2 \times 2\theta$)

$\displaystyle = 2\sin 2\theta\cos 2\theta$

$\displaystyle = 2\times 2\sin\theta\cos\theta \times (2\cos^2\theta - 1)$

I'll leave you to simplify this.

Originally Posted by Backlock
13) Prove the following identitys:

a) $\displaystyle cot^2 x+1 = cosec^2 x$
Take $\displaystyle \cos^2 x + \sin^2x=1$ and divide both sides by $\displaystyle \sin^2x$:

$\displaystyle \frac{\cos^2x}{\sin^2x} + 1 = \frac{1}{\sin^2x}$

$\displaystyle \Rightarrow$ ...?

Originally Posted by Backlock
b) $\displaystyle (tan x+sec x)/(sec x-cos x+tan x) = cosec x$
Note that $\displaystyle \tan x = \frac{\sin x}{\cos x}$ and $\displaystyle \sec x = \frac{1}{\cos x}$ and multiply top and bottom of $\displaystyle \frac{\tan x + \sec x}{\sec x - \cos x + \tan x}$ by $\displaystyle \cos x$ to get:

$\displaystyle \frac{\tan x + \sec x}{\sec x - \cos x + \tan x} = \frac{\sin x + 1}{1 - \cos^2x + \sin x}$

$\displaystyle = \frac{\sin x + 1}{\sin^2x + \sin x}$

Can you finish it now? (Take out a common factor in the denominator.)

Originally Posted by Backlock
c) $\displaystyle sin^3 x-cos^3 x = (1+sin x*cos x)(sin x-cos x)$

For c) we were told to go about doing the right hand side first.

Thanks again!
OK, but I think it's easier to start with the LHS, noting that $\displaystyle a^3 - b^3$ factorises to $\displaystyle (a-b)(a^2 + ab + b^2)$. So:

$\displaystyle \sin^3x - \cos^3 x= (\sin x - \cos x) (\sin^2x + \sin x \cos x + \cos^2x)$

$\displaystyle = (\sin x - \cos x)(1 + \sin x \cos x)$, using $\displaystyle \sin^2x + \cos^2x = 1$