csc x - sin x = cot x * cos x

[screambloodygore]

This is supposed to be solved using just the eight trigonometric identities (3 reciprocal identities, tangent and cotangent identities, and the 3 Pythagorean identities): everything else up to this point, I've been able to solve, but this question and the next 20 questions in the exercise are all Swahili to me-- I have no idea where to start on any of them. I've gone as far as this for the equation in the thread title:

left-hand side: csc x - sin x
= 1 / (sin x) - sin x

I'm not even sure where to go after this- the book doesn't give me answers in the back for the trigonometric identities- it says I should "ask my teacher," but I'm doing the class alone and sending the assignments into a government office to be marked.

There's another equation that looks like it'll be solved in almost the exact same way as the first one:

sec x - cos x = tan x * sin x

...er, please help? I'm totally lost on this stuff. I've read and taken notes on the next section, "Strategies for Proving Identities," but I can't seem to apply any of them here or I'm just really tired and frustrated.

 

[galactus]

This is supposed to be solved using just the eight trigonometric identities (3 reciprocal identities, tangent and cotangent identities, and the 3 Pythagorean identities): everything else up to this point, I've been able to solve, but this question and the next 20 questions in the exercise are all Swahili to me-- I have no idea where to start on any of them. I've gone as far as this for the equation in the thread title:

left-hand side: csc x - sin x
= 1 / (sin x) - sin x

You're on the right track.

\(\displaystyle \L\\csc(x)-sin(x)=cot(x)cos(x)\)

LHS:

\(\displaystyle \L\\=\frac{1}{sin(x)}-sin(x)\)

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

\(\displaystyle \L\\=\frac{cos^{2}(x)}{sin(x)}\)

\(\displaystyle \L\\=\frac{cos(x)}{sin(x)}cos(x)\)

\(\displaystyle \L\\=cot(x)cos(x)\)

 

[soroban]

Hello, screambloodygore!

Now that galactus has shown you how to verify the identity,
\(\displaystyle \;\;\)you may be inspired to experiment . . . or not.

Try the second problem from the right side:

\(\displaystyle \L\;\;\;\sec x\,-\,\cos x \:= \:\tan x\,\cdot\,\sin x\)


We have: \(\displaystyle \L\,\tan x\,\cdot\,\sin x\;=\;\frac{\sin x}{\cos x}\,\cdot\,\sin x\;=\;\frac{\sin^2x}{\cos x} \;=\;\frac{1\,-\,\cos^2x}{\cos x}\)

Make two fractions: \(\displaystyle \L\,\frac{1}{\cos x}\,-\,\frac{\cos^2x}{\cos x}\;=\;\sec x\,-\,\cos x\)

 

[screambloodygore]

Ah, thank you. I tried a whole bunch of other solutions on 2 pieces of paper and got as far as: (1 - sin^2 x) / sin x
= (cos^2 x) / sin x

...but after that, I didn't know what to do.
This will be really helpful for solving the rest of the problems in the section, because I think this is almost the exact technique they require.

Thanks dudes.