trig help?

[jordan83]

sin(theta)+cos(theta)/cos(theta) - sin(theta)-cos(theta)/sin(theta) = sec(theta)csc(theta)

I need help proving these are equal. Thanks!

 

[stapel]

I need help proving these are equal.

So this is an identity, and the instructions are to "prove" the statement?

You have posted:

. . . . .\(\displaystyle \Large{\sin{(\theta)}\mbox{ }+\mbox{ }\frac{\cos{(\theta)}}{\cos{(\theta)}}\mbox{ }-\mbox{ }\sin{(\theta)}\mbox{ }-\mbox{ }\frac{\cos{(\theta)}}{\sin{(\theta)}}\mbox{ }=\mbox{ }\sec{(\theta)}\csc{(\theta)}}\)

If this is what you meant, then try evaluating at (pi)/2: this "identity" isn't true ever.

Note: Your equation simplifies to:

. . . . .\(\displaystyle \large{1\mbox{ }-\mbox{ }\cot{(\theta)}\mbox{ }=\mbox{ }\sec{(\theta)}\csc{(\theta)}}\)

If you meant something else by your equation, please reply with clarification, including the instructions and showing what you have tried thus far.

Thank you.

Eliz.

 

[jordan83]

sorry its (sin(theta) + cos(theta)) all over cos(theta) - (sin(theta)-cos(theta)) all over sin(theta) = sec(theta)csc(theta)

I cancelled the cos and sins on them but that got me no where. I need some help. Thank you.

 

[stapel]

sorry its (sin(theta) + cos(theta)) all over cos(theta) - (sin(theta) - cos(theta)) all over sin(theta) = sec(theta)csc(theta)

Do you mean the "minus" sign, highlighted above, to split this into two fractions, or is this, as you've posted, one really big fraction with lots of fractions inside?

I cancelled the cos and sins on them but that got me no where.

I'm sorry, but I don't know what you mean by this. Please reply showing your steps.

Thank you.

Eliz.

 

[jordan83]

The minus sign is splitting it up.

I cancelled the sin and cos from the like terms of the numerator and denominator and then I am lost. Thanks.

 

[stapel]

The minus sign is splitting it up.

So we have:


. . . . .\(\displaystyle \Large{\frac{\sin{(\theta)}\mbox{ } + \mbox{ }\cos{(\theta)}}{\cos{(\theta)}}\mbox{ }- \mbox{ }\frac{\sin{(\theta)}\mbox{ } - \mbox{ }\cos{(\theta)}}{\sin{(\theta)}}\mbox{ } = \mbox{ }\frac{\sec{(\theta)}}{\csc{(\theta)}}}\)


Converting to a common denominator (usually a good first step when faced with fractions), we get:


. . . . .\(\displaystyle \Large{\frac{\sin{(\theta)}\cos{(\theta)}\mbox{ }+\mbox{ }\cos^2{(\theta)}\mbox{ }-\mbox{ }\sin{(\theta)}\cos{(\theta)}\mbox{ }+\mbox{ }\cos^2{(\theta)}}{\cos{(\theta)}\sin{(\theta)}}\mbox{ }=\mbox{ }\frac{\sec{(\theta)}}{\csc{(\theta)}}}\)


I'd simplify and see where I could from there.

I cancelled the sin and cos from the like terms of the numerator and denominator and then I am lost. Thanks.

Terms don't cancel, of course -- only factors do -- so I'm guessing you mean that you cancelled the common factors.

But I'm afraid I'm not seeing these common factors in the original equation, so you must be working this in a very different way.

Please reply showing your steps. Thank you.

Eliz.

 

[soroban]

Hello, jordan83!

I <u>think</u> I know what you meant . . . but please learn to use parentheses!

\(\displaystyle \L\frac{\sin\theta\,+\,\cos\theta}{\cos\theta}\,-\,\frac{\sin\theta\,-\,\cos\theta}{\sin\theta}\;=\;\sec\theta\cdot\csc\theta\)

On the left side, get a common denominator: . \(\displaystyle \L\frac{\sin\theta}{\sin\theta}\cdot\frac{\sin\theta\,+\,\cos\theta}{\cos\theta}\,-\,\frac{\cos\theta}{\cos\theta}\cdot\frac{\sin\theta\,-\,\cos\theta}{\sin\theta}\)


. . \(\displaystyle \L=\;\frac{\sin\theta\cdot(\sin\theta\,+\,\cos\theta)\,-\,\cos\theta\cdot(\sin\theta\,-\,\cos\theta)}{\cos\theta\cdot\sin\theta}\;=\;\frac{\sin^2\theta\,+\,\sin\theta\cdot\cos\theta\,-\,\sin\theta\cdot\cos\theta\,+\,\cos^2\theta}{\cos\theta\cdot\sin\theta}\)


. . \(\displaystyle \L=\;\frac{\sin^2\theta\,+\,\cos^2\theta}{\cos\theta\cdot\sin\theta}\;=\;\frac{1}{\cos\theta\cdot\sin\theta}\;=\;\sec\theta\cdot\csc\theta\)