# Trig simplification issues

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• Oct 22nd 2010, 09:07 PM
abzolutezero
Trig simplification issues
I'v been sick in my college math class and have been a bit behind. I really could use some help and have been slamming my head against my text book trying to figure some of these simplifications out.(Headbang) One in particular that's been driving me nuts is this.

cosx/(((cosx-1)/(cosx))-1) Sorry for my overuse of parentheses here not sure how to make it look all professional.

another is
if Asin(squared)x+Bcos(squared)x =1 show that
sin(squared)x=(1-b)/a-b) and tan(squared)x=(b-1)/(1-a)

my last one is
suppose Asinx+cosx=1 and bsinx-cosx=1 show that AB=1

any help you can offer would be much appreciated as I've been working on these problems for hours and solved nothing.
• Oct 22nd 2010, 10:30 PM
Soroban
Hello, abzolutezero!

Here's the third one . . .

Quote:

$\text{Suppose: }\:\begin{Bmatrix}A\sin x+\cos x &=& 1 & [1] \\ B\sin x-\cos x &=& 1 & [2] \end{Bmatrix}$

$\text{Show that: }\:AB\:=\:1$

From [1], we have: . $A \;=\;\dfrac{1-\cos x}{\sin x}\;\;[3]$

From [2], we have: . $B \;=\;\dfrac{1+\cos x}{\sin x}\;\;[4]$

Multiply [3] and [4]:

. . $\displaystyle A\cdot B \;=\;\frac{1-\cos x}{\sin x}\cdot\frac{1+\cos x}{\sin x} \;=\; \frac{1-\cos^2\!x}{\sin^2\!x} \;=\;\frac{\sin^2\!x}{\sin^2\!x} \;=\;1$
• Oct 23rd 2010, 04:22 AM
skeeter
Quote:

Originally Posted by abzolutezero
cosx/(((cosx-1)/(cosx))-1)

if Asin(squared)x+Bcos(squared)x =1 show that
sin(squared)x=(1-b)/a-b) and tan(squared)x=(b-1)/(1-a)

if this is what you meant for the fist problem ...

$\displaystyle \frac{\cos{x}}{\frac{\cos{x}-1}{\cos{x}}-1}$

... multiply numerator and denominator by $\cos{x}$ , then simplify.

for the second problem(s), I assume the cap A's and B's are the same as the lower case ones ...

$a\sin^2{x} + b\cos^2{x} = 1$

$a\sin^2{x} + b(1-\sin^2{x}) = 1$

distribute $b$ , then solve for $\sin^2{x}$.

$a\sin^2{x} + b\cos^2{x} = 1$

$a\sin^2{x} + b\cos^2{x} = \sin^2{x} + \cos^2{x}
$

$a\sin^2{x} - \sin^2{x} = \cos^2{x} - b\cos^2{x}
$

$\sin^2{x}(a - 1) = \cos^2{x}(1 - b)$

$\frac{\sin^2{x}}{\cos^2{x}} = \frac{1-b}{a-1}
$

$\tan^2{x} = \frac{1-b}{a-1} = \frac{b-1}{1-a}$