Verify Identy: {(sin^4x-sin^2x)/(1-sec^2x)cos^4x} = ....

[messa]

Can you tell me if I'm doing this correct, and if so, where can I go from here?

Verify the identity: {(sin^4x-sin^2x)/(1-sec^2x)cos^4x} = {(cos^3x+1/2sin2xsinx)/(secx)} + {(sin^3x+1/2sin2xcosx)/cscx}

I started working in the right side which I'm now regretting, but this is what I have:

{(cos^3x+1/4cosx-1/4cos3x)/secx} + {(sin^3x+1/4sin3x+1/4sinx)/cscx} = {(1/4cos^4x-1/4cox3x)/secx} + {(1/4sin^4x+1/4sin3x)/cscx}

What a mess!

 

[soroban]

Hello, messa!

I had to do some Olympic-level gymnastics, but here it is . . .

Verify the identity:
. . \(\displaystyle \L\frac{\sin^4x\,-\,\sin^2x}{(1\,-\,\sec^2x)\cos^4x} \:= \:\frac{\cos^3x\,+\,\frac{1}{2}\cdot\sin2x\cdot\sin x}{\sec x}\,+\,\frac{\sin^3x\,+\,\frac{1}{2}\cdot\sin2x\cdot\cos x}{\csc x}\)

The left side is: \(\displaystyle \L\:\frac{\sin^2x(\sin^2x\,-\,1)}{\left(1\,-\,\frac{1}{\cos^2x}\right)\cos^4x} \;=\;\frac{-\sin^2x\cdot(1\,-\,\sin^2x)}{\cos^4x\,-\,\cos^2x}\;=\;\frac{-\sin^2x\cdot\sout{\cos^2x}}{\sout{\cos^2x}(\cos^2x\,-\,1)}\)

. . \(\displaystyle \L=\;\frac{-\sin^2x}{\cos^2x\,-\,1} \;=\;\frac{-\sin^2x}{-(1\,-\,\cos^2x)} \;=\;\frac{-\sin^2x}{-\sin^2x} \;=\;1\)


Take a deep breath . . . here we go!

. . \(\displaystyle \L1\;=\;\cos^2x\,+\,\sin^2x \;=\;\frac{\cos x}{\frac{1}{\cos x}} \,+\, \frac{\sin x}{\frac{1}{\sin x}} \;=\;\frac{\cos x}{\sec x}\,+\,\frac{\sin x}{\csc x}\)

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

. . \(\displaystyle \L=\;\frac{\cos^3x\,+\,\sin^2x\cdot\cos x}{\sec x} \,+\,\frac{\sin^3x\,+\,\sin x\cdot\cos^2x}{\csc x}\)

. . \(\displaystyle \L=\;\frac{\cos^3x\,+\,(\sin x\cdot\cos x)\sin x}{\sec x} \,+\,\frac{\sin^3x\,+\,(\sin x\cdot\cos x)\cos x}{\csc x}\)

. . \(\displaystyle \L=\;\frac{\cos^3x\,+\,\frac{1}{2}\cdot\sin2x\cdot\sin x}{\sec x}\,+\,\frac{\sin^3x\,+\,\frac{1}{2}\cdot\sin2x\cdot\cos x}{\csc x}\;\) . . . ta-DAA!

 

[messa]

Holy cow! Thank you so much. I makes complete sense when I see it, but I have such a difficult time piecing everything together! I never would have thought to make it equal 1 and then go from there. Thanks!