# Equations and results involving double angle formulae

• February 28th 2010, 11:48 PM
RAz
Equations and results involving double angle formulae
1. I would like to know if this is correct:

$5sinx-2secx=0$ where x is an element of 0 and 360 (2pi in radians)

$5sinx-\frac {2}{cosx}=0$

$5sinxcosx-2=0$

$sin5x-2=0$

$\therefore sinx=\frac {2}{3}, \frac {2}{3} + \pi, \frac {2}{3} + 2\pi$

2. I would like to know where to continue in this question:

$sin3xcosx-cos3xsinx=\frac {\sqrt{3}}{2}$

$3sinx-4cos^3xcosx-4cos^3x-3cosx=$

$3sinxcosx-4cos^4x-4sincos^3x-3cos^2xsinx=$

...I don't really know where to go from here and how to find angles for x.

3. I am being introduced to product and sum formulas and I don't know how to tackle the questions.

eg1. $sinx=\frac{3}{2}, 0

Find the exact value of sin2x, cos2x.

The extra formulas given for sinx and cosx are:

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

and

$cosx=\frac {1-tan^2 \frac{x}{2}}{1+tan^2\frac{x}{2}}$

I know all the normal ones like sin2x=2sinxcosx etc., just not how to apply them.

If you could help me out here that would be great :D
• March 1st 2010, 03:57 AM
Hello RAz

red.
Quote:

Originally Posted by RAz
1. I would like to know if this is correct:

$5sinx-2secx=0$ where x is an element of 0 and 360 (2pi in radians)

$5sinx-\frac {2}{cosx}=0$

$5sinxcosx-2=0$

$sin5x-2=0$ No. 5 sin x cos x = 2.5 sin 2x.

$\therefore sinx=\frac {2}{3}, \frac {2}{3} + \pi, \frac {2}{3} + 2\pi$

2. I would like to know where to continue in this question:

$sin3xcosx-cos3xsinx=\frac {\sqrt{3}}{2}$

$3sinx-4cos^3xcosx-4cos^3x-3cosx=$ Don't do it like this. Use sin 3x cos x - cos 3x sin x = sin (3x-x) = sin 2x

$3sinxcosx-4cos^4x-4sincos^3x-3cos^2xsinx=$

...I don't really know where to go from here and how to find angles for x.

3. I am being introduced to product and sum formulas and I don't know how to tackle the questions.

eg1. $sinx=\frac{3}{2}, 0 This doesn't make sense. Do you mean tan x = 3/2?

Find the exact value of sin2x, cos2x.

The extra formulas given for sinx and cosx are:

$sinx=\frac{2\color{red}\tan\color{black}\frac{x}{2 }}{1+tan^2 \frac{x}{2}}$

and

$cosx=\frac {1-tan^2 \frac{x}{2}}{1+tan^2\frac{x}{2}}$

I know all the normal ones like sin2x=2sinxcosx etc., just not how to apply them. Don't use them here. Evaluate sin 2x and cos 2x, using the formula above, but substituting 2x in place of x.

If you could help me out here that would be great :D

Can you try these again now?