Fundamental Identities: sin^2(x) (csc^2(x) - 1)

[azdressagenut]

I am currently working on fundamental Identities and I am struggling with this topic. This question I have been attempting for a while now and everything I have tried is unsuccessful.
Directions: Write each expression in terms of sine and cosine, and simplify so that no quotients appear in the final expression.
sin[sup:1ns94gm6]2[/sup:1ns94gm6]x(csc[sup:1ns94gm6]2[/sup:1ns94gm6]x - 1)

 

[Loren]

Re: Fundamental Identities

Change the csc[sup:2cdu6hni]2[/sup:2cdu6hni]x to 1/sin[sup:2cdu6hni]2[/sup:2cdu6hni]x then combine what's inside the parenthesis. You should see the next step.

 

[stapel]

sin[sup:24mhird4]2[/sup:24mhird4]x(csc[sup:24mhird4]2[/sup:24mhird4]x - 1)

To expand upon the (very good) suggestion in the previous reply, you will want to convert to sines and cosines (just sines, in this case), get a common denominator, and then see what you can do from there.

. . . . .\(\displaystyle \sin^2(x)\left(\frac{1}{\sin^2(x)}\, -\, 1\right)\)

. . . . .\(\displaystyle \sin^2(x)\left(\frac{1}{\sin^2(x)}\, -\, \frac{\sin^2(x)}{\sin^2(x)}\right)\)

. . . . .\(\displaystyle \frac{\sin^2(x)}{1}\,\left(\frac{1\, -\, \sin^2(x)}{\sin^2(x)}\right)\)

Do the obvious simplification, and then apply the basic Pythagorean identity to simplify what's left. :wink:

 

[soroban]

Re: Fundamental Identities

Hello, azdressagenut!

Write each expression in terms of sine and cosine, and simplify
so that no quotients appear in the final expression.

. . \(\displaystyle \sin^2\!x \left(\csc^2\!x - 1\right)\)

\(\displaystyle \text{We have: }\;\sin^2\!x\underbrace{\left(\csc^2\!x - 1\right)}_{\text{This is }\cot^2\!x} \;=\;\sin^2\!x\cdot\cot^2\!x \;=\;\sin^2\!x\cdot\frac{\cos^2\!x}{\sin^2\!x} \;=\;\cos^2\!x\)