Trigonometric Identities Help!!

So I started AS maths last week and I'm having some trouble with identities. Please please please can someone try to help me!

1. (1/sinx) - tanxcosx = cos^2x/sinx
2. tanxsinx/1-cosx = 1 + 1/cosx
3. (cosx+sinx)^2 + (cosx-sinx)^2 =2
4. sinxtanx + cosx = 1/cosx
5. 1/sinx - 1/tanx = 1 - cosx/sinx

Any help would be greatly appreciated!

What are you stuck on exactly? At this point you should know that:

$\displaystyle \sin^2x+\cos^2x\equiv 1$

$\displaystyle \frac{\sin{x}}{\cos{x}}\equiv \tan{x}$

and work from these.

Yes I found those in the textbook but I don't understand how to rearrange them to simplify those equations

I'll do the first one for you as an example:

First rewrite $\tan{x}$ as $\displaystyle \frac{\sin{x}}{\cos{x}}$. You can see that multiplying this by $\cos{x}$ cancels out the denominator right away.

You are left with: $\displaystyle \frac{1}{\sin{x}}-\sin{x}=\frac{\cos^2x}{\sin{x}}$

Now multiply through by sin(x) to cancel the 1/sinx and you are left with: $\displaystyle 1-\sin^2x=\cos^2x$ which is a true identity.

Do the similar procedure for the other ones.

I'll do the first one for you as an example:

First rewrite $\tan{x}$ as $\displaystyle \frac{\sin{x}}{\cos{x}}$. You can see that multiplying this by $\cos{x}$ cancels out the denominator right away.

You are left with: $\displaystyle \frac{1}{\sin{x}}-\sin{x}=\frac{\cos^2x}{\sin{x}}$

Now multiply through by sin(x) to cancel the 1/sinx and you are left with: $\displaystyle 1-\sin^2x=\cos^2x$ which is a true identity.

Do the similar procedure for the other ones.

Oh my goodness you are a genius! I'll have a go at the others and see how I get on. Thank you!!

It is very bad practice to assume that an identity is true and manipulate it to arrive at something true.

If you do this you have to be sure that every step you make is reversible which isn't always obvious.

With trig identities you should always start from one side and show that the other side is true.

Right I'm stuck again. On number 3, cos^2x + sin^2 simplifies to 1, but how do I simplify cos^2x - sin^2x?

Please do not do what RDKGames has done for trig identities.

You should always start from one side of a trig identity and aim to arrive at the other side. Never manipulate the whole identity like you would an equation.

The question didn't really make it clear and I assumed they want you to prove whether something turns out to be a true identity or not.

Show your working, it shouldn't simplify down to a minus. Also as the person above said, leave the right hand side along and work with left one.

That doesn't make sense.

I'm busy now but if you have time can you please show the OP how to do question 1 using the correct method?

Oh yes I see now. (-sin^2x)^2 is positive, not negative. Thanks again

To me it does, but okay.



Q1 corrected:

Rewrite tanx as sinx over cosx, the cosx cancels with the denominator as before. The RHS wants the answer to be as something over sinx, so you need to combine the LHS into a single fraction under the same denominator. The first term already has sinx as denominator, we can leave it. The second term, sinx, can be turned into a fraction with denominator sinx by:

$sinx \cdot 1 = sinx \cdot \frac{sinx}{sinx} = \frac{sin^2x}{sinx}$

Now add the two fractions and you get: $\frac{1-sin^2x}{sinx}=\frac{cos^2x}{sinx}$ at which point we need to show that the numerator is equal to $cos^2x$ and indeed this is true from rearranging $sin^2x+cos^2x=1$

Oh right I see now! So would the answer be cos^2x/sinx = cos^2x/sinx??

To me it does, but okay.

Exactly. With these what you are essentially doing is turning the left hand side of the identity into the right hand side.

So I've moved on to question 4 now. I've turned sinx into tanx/cosx and tanx into sinx/cosx. Then I've combined the two fractions to make tanxsinx/cosx. Now I'm stuck because I can't just cross out the cosx's. Please please please can someone help!

So I've moved on to question 4 now. I've turned sinx into tanx/cosx and tanx into sinx/cosx. Then I've combined the two fractions to make tanxsinx/cosx. Now I'm stuck because I can't just cross out the cosx's. Please please please can someone help!