# Write the expressions in terms of cosine?

• March 11th 2009, 10:09 AM
mj.alawami
Write the expressions in terms of cosine?
Q)Write the following expression in terms of cosine both in a simplified form?

$\frac{1-csc^2x}{csc^2x}$

Attempt- Is my attempt Correct (Wondering))

$-1(sin^2x-1) =-cos^2x$

Thank you
• March 11th 2009, 10:19 AM
e^(i*pi)
Quote:

Originally Posted by mj.alawami
Q)Write the following expression in terms of cosine both in a simplified form?

$\frac{1-csc^2x}{csc^2x}$

Attempt- Is my attempt Correct (Wondering))

$-1(sin^2x-1) =-cos^2x$

Thank you

$csc^2(x) = \frac{1}{sin^2(x)}$

$
1- \frac{1}{sin^2(x)} = \frac{sin^2(x) - 1}{sin^2(x)}$

Overall the sum becomes:

$\frac{sin^2(x) - 1}{sin^2(x)csc(x)}$ and so the denominator will be sin(x)

$sin^2(x) - 1 = -cos^2(x) and sin(x) = \sqrt{1-cos^2(x)}$

therefore we have $\frac{-cos^2(x)}{\sqrt{1-cos^2(x)}}$

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From your expression you can say that $-1(sin^2(x)-1) = 1-sin^2(x) = cos^2(x)$
• March 11th 2009, 10:26 AM
mj.alawami
Quote:

Originally Posted by e^(i*pi)
$csc^2(x) = \frac{1}{sin^2(x)}$

$
1- \frac{1}{sin^2(x)} = \frac{sin^2(x) - 1}{sin^2(x)}$

Overall the sum becomes:

$\frac{sin^2(x) - 1}{sin^2(x)csc(x)}$ and so the denominator will be sin(x)

$sin^2(x) - 1 = -cos^2(x) and sin(x) = \sqrt{1-cos^2(x)}$

therefore we have $\frac{-cos^2(x)}{\sqrt{1-cos^2(x)}}$

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From your expression you can say that $-1(sin^2(x)-1) = 1-sin^2(x) = cos^2(x)$

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Your equation is correct but you can simplify the $csc^2$ in the denominator to $\frac{1}{sin^2}$ and then you can cancel the numerator and get $sin^2-1$

But i am not sure of my attempt

Thank you
• March 11th 2009, 11:27 AM
e^(i*pi)
Quote:

Originally Posted by mj.alawami
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Your equation is correct but you can simplify the $csc^2$ in the denominator to $\frac{1}{sin^2}$ and then you can cancel the numerator and get $sin^2-1$

But i am not sure of my attempt

Thank you

My apologies I misread the question. Yes you can cancel the denominator because $sin^2(x)csc^2(x) = 1$