Thread: Need help with trig identities

If you could go step by step it would be helpful!

cotx+tanx = secxcscx



sin^4x - cos^4x = 2sin^2x - 1



cot^2x-1/1+cot^2x = 1-2sin^2x
Thanks!

Originally Posted by Thatoneguy12345

cot^2x-1/1+cot^2x = 1-2sin^2

Wait, how do you get the -1+1-1 on the 2nd step?

Originally Posted by Thatoneguy12345

Wait, how do you get the -1+1-1 on the 2nd step?

1-1 is zero right? If you add it to the equation it doesn't change the value of the equation. But you can use it to simplify the equation, like I did.

For example if you have the equation

5+2 = 7

That's the same as

5+2+(1-1)=7

Oh whoa, thanks! Never knew you could do it that way!

Q1
Write LHS in terms of sin and cos
Get a common denominator and add
You should recognise and simplify the top
Split up the product
You should now have the RHS
TRY IT!!

Q2 Hint: a^4 - b^4 =(a^2 + b^2)(a^2 - b^2)
The first bracket will be =1
Look at the RHS (there is no cos in sight!)
So, replace (cos x)^2 with 1 - (sin x)^2
Too easy!

Q3 Use the identity 1 + (cot x)^2 = (cosec x)^2 on the bottom
Dividing by (cosec x)^2 is the same as multiplying by (sin x)^2
So now you should have (sin x)^2 x ((cot x)^2 -1)
Write cot^2 in terms of sin and cos
Expand out bracket
Do the same thing you did at the end of Q2 to get rid of cos^2
Done

Originally Posted by Thatoneguy12345

cotx+tanx = secxcscx

You choose the common denominator of and multiply through. You get:

Use the identity




Edit: If you still need me to show you the second one after Debsta explained it, post the request. Debsta did a good job of explaining it.

My thoughts exactly!

Thanks a lot you two!