# help correcting my trig equations and proving one

• May 26th 2009, 08:27 AM
llkkjj24
help correcting my trig equations and proving one
hi, i have this HW for my class assessment mark so i really want to get it all perfect, its some tedious work, but help is greatly appreciated.Thanks in advance. pls help me check if i made any mistakes in my calculations

1) solve 5sinX - 2cos^2X - 1 = 0 range(for 0 <= x <= 2pie) <= is great equal sign

i worked out and got sinX = 1/2 and sinX = -3

for sin = 1/2 converting to degress(using calculator inverse sin) is 30,
using quadrant diagram x1 = 30 degrees
x2 = 180-30 = 150
sinx = -3 is undefined? though i don't know if i did any mistake in my previous calculations

btw if the range is given in terms of pie, does the teacher expects the answer to be given in terms of pie?, i can't figure out what 150 is in terms of pie(Rofl)

2) sec^2X - 4tanX = 0 for (0<= x <= 180)

using identities sec^2X = tan^2X +1 thus
tan^2 X - 4tan X +1 = 0
since it can't be factored mentally , i used the quadratic equation to factor

got tan X = 3.732

using calculator inverse tan and quadrant diagram got
X1 = 75
x2 = 180 + 75 = 255
(all in terms of degees)

next tanX = 0.268
using above methods got X1= 15.5
X2 = 180 - 15.5 = 164.5

3) 3sinX = 5sinXcosX (for -90 <x<90)
completely lost, tried diveding everything by cosX, didn't work
and where do i start of finish in the quadrant diagram if its -90

4) 3 sinX = 5sinXcosX (for -90<x<90)

got 3/5 = cosx
inverse of cos 3/5 = 53.1(degrees)

stuck on the quadrant diagram, becasue of the range.

help is greatly appreciated, i tried, just got stuck on some, and its a very important piece of HW
• May 26th 2009, 08:29 AM
llkkjj24
last one,how do you prove this ?

(cotX + cosecX)^2 = 1+cosX/1-cosX

some people are just born to fail maths
• May 26th 2009, 08:58 AM
masters
Quote:

Originally Posted by llkkjj24
last one,how do you prove this ?

(cotX + cosecX)^2 = 1+cosX/1-cosX

some people are just born to fail maths

Hi llkkjj24,

I'll jump on this one. Working with left side, only:

$\displaystyle (\cot x+\csc x)^2=\frac{1+\cos x}{1-\cos x}$

$\displaystyle \left(\frac{\cos x}{\sin x}+\frac{1}{\sin x}\right)^2=$

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

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

$\displaystyle \frac{(\cos x+1)(\cos x+1)}{1-\cos^2 x}=$

$\displaystyle \frac{(\cos x+1)(\cos x+1)}{(1-\cos x)(1+\cos x)}=$

$\displaystyle {\color{red}\frac{1+\cos x}{1-\cos x}}$
• May 26th 2009, 09:26 AM
masters
Quote:

Originally Posted by llkkjj24
hi, i have this HW for my class assessment mark so i really want to get it all perfect, its some tedious work, but help is greatly appreciated.Thanks in advance. pls help me check if i made any mistakes in my calculations

1) solve 5sinX - 2cos^2X - 1 = 0 range(for 0 <= x <= 2pie) <= is great equal sign

i worked out and got sinX = 1/2 and sinX = -3

for sin = 1/2 converting to degress(using calculator inverse sin) is 30,
using quadrant diagram x1 = 30 degrees
x2 = 180-30 = 150
sinx = -3 is undefined? though i don't know if i did any mistake in my previous calculations

btw if the range is given in terms of pie, does the teacher expects the answer to be given in terms of pie?, i can't figure out what 150 is in terms of pie(Rofl)

$\displaystyle 30^{\circ}=\frac{30}{1}\cdot \frac{\pi}{180}=\frac{\pi}{6}$ radians.

$\displaystyle 150^{\circ}=\frac{150}{1}\cdot \frac{\pi}{180}=\frac{5\pi}{6}$ radians.
• May 26th 2009, 09:58 AM
masters
Quote:

Originally Posted by llkkjj24
3) 3sinX = 5sinXcosX (for -90 <x<90)
completely lost, tried diveding everything by cosX, didn't work
and where do i start of finish in the quadrant diagram if its -90

4) 3 sinX = 5sinXcosX (for -90<x<90)

got 3/5 = cosx
inverse of cos 3/5 = 53.1(degrees)

stuck on the quadrant diagram, becasue of the range.

help is greatly appreciated, i tried, just got stuck on some, and its a very important piece of HW

3) and 4) appear to be the same problem.

-53.1 works and is within the range.
• May 26th 2009, 10:16 AM
llkkjj24
thank you very much for the help(Clapping)
• May 26th 2009, 01:26 PM
Soroban
Hello, llkkjj24!

Another approach . . .

Quote:

Prove: .$\displaystyle (\cot x + \csc x)^2 \:=\:\frac{1+\cos x}{1-\cos x}$

Multiply the right side by $\displaystyle \frac{1+\cos x}{1 +\cos x}$

. . $\displaystyle \frac{1+\cos x}{1-\cos x}\cdot{\color{blue}\frac{1+\cos x}{1 + \cos x}} \;=\;\frac{(1+\cos x)^2}{1-\cos^2\!x} \;=\;\frac{(1+\cos x)^2}{\sin^2\!x}$

. . . $\displaystyle = \;\left(\frac{1+\cos x}{\sin x}\right)^2 \;= \;\left(\frac{1}{\sin x} + \frac{\cos x}{\sin x}\right)^2 \;=\;(\csc x + \cot x)^2$