# 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 $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ with $a = \cos^2 x$ and $b = \sin^2 x$

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
$\cos x = \frac{a^2 - m^2}{a^2 + m^2}$
cos x is also adjacent over hypotenuse in a right triangle, so let's say that the adjacent side is $a^2 - m^2$ and that the hypotenuse is $a^2 + m^2$.

Use the Pythagorean theorem to find the remaining (opposite) side:
\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 $0 < x < \frac{\pi}{2}$, we can drop the $\pm$ sign. y = 2am

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

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

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

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

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