# Thread: trigonometry equatioonss partt 2 =(

1. ## trigonometry equatioonss partt 2 =(

hii guyzz, having trouble with MORE of these trig equations ><""

use the factors of xcubed - ycubed to show that
cos^6x - sin^6x = (1- 1/4sin^2x)cos2x

by the way, this ^means to the power to, and its just to the power of 6, and NOT to the power of 6x if anyone gets confued, the x is like theta

AND

if cosx = (a^2 - m^2)/(a^2 + m^2) and 0<x<pi/2 express tanx and sin2x in terms of a and m

thankyou sooooo much to anyone who can give me some help in these equations, i'm so grateful =D

2. Originally Posted by iiharthero
hii guyzz, having trouble with MORE of these trig equations ><""

use the factors of xcubed - ycubed to show that
cos^6x - sin^6x = (1- 1/4sin^2x)cos2x

by the way, this ^means to the power to, and its just to the power of 6, and NOT to the power of 6x if anyone gets confued, the x is like theta
Hi

You are told to use $\displaystyle a^3 - b^3 = (a-b)(a^2+ab+b^2)$ with $\displaystyle a = \cos^2 x$ and $\displaystyle b = \sin^2 x$

3. Originally Posted by iiharthero
if cosx = (a^2 - m^2)/(a^2 + m^2) and 0<x<pi/2 express tanx and sin2x in terms of a and m
$\displaystyle \cos x = \frac{a^2 - m^2}{a^2 + m^2}$
cos x is also adjacent over hypotenuse in a right triangle, so let's say that the adjacent side is $\displaystyle a^2 - m^2$ and that the hypotenuse is $\displaystyle a^2 + m^2$.

Use the Pythagorean theorem to find the remaining (opposite) side:
\displaystyle \begin{aligned} (a^2 - m^2)^2 + y^2 &= (a^2 + m^2)^2 \\ a^4 - 2a^2m^2 + m^4 + y^2 &= a^4 + 2a^2m^2 + m^4 \\ -2a^2m^2 + y^2 &= 2a^2m^2 \\ y^2 &= 4a^2m^2 \\ y &= \pm 2am \end{aligned}

Since $\displaystyle 0 < x < \frac{\pi}{2}$, we can drop the $\displaystyle \pm$ sign. y = 2am

So,
adjacent side is $\displaystyle a^2 - m^2$,
opposite side is $\displaystyle 2am$, and
hypotenuse is $\displaystyle a^2 + m^2$.

Use $\displaystyle \sin x = \frac{\text{opp}}{\text{hyp}}$ to find sin x.

Use $\displaystyle \tan x = \frac{\text{opp}}{\text{adj}}$ or $\displaystyle \frac{\sin x}{\cos x}$ to find tan x.

Use $\displaystyle \sin 2x = 2\sin x \cos x$ to find sin 2x.

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