Trig identity: sin^4 theta + cot^2 theta = (1/4)sin^2 2theta

[hplovr411]

I need to prove:

sin^4 THETA + cot^2 THETA = (1/4)sin^2 (2THETA)

Anyone who can solve the problem step by step would be greatly appreciated!
Thanks!

 

[galactus]

\(\displaystyle \L\\sin^{4}({\theta})+cot^{2}({\theta})=\frac{sin^{2}(2{\theta})}{4}\)

I may be wrong, but I don't believe that's an identity.

Are you sure you copied it correctly?.

 

[hplovr411]

Yep thats right. The teacher comes up with really weird problems. We have to do this on our final

heres how the teacher typed it: SIN^4 THEDA COT^2 THEDA = 1/4 SIN ^2 (2) THEDA

 

[pka]

\(\displaystyle \L\begin{array}{rcl}
\sin ^4 (t)\cot ^2 (t) & = & \sin ^4 (t)\frac{{\cos ^2 (t)}}{{\sin ^2 (t)}} \\
& = & \sin ^2 (t)\cos ^2 (t) \\
& = & \left( {\frac{1}{2}\sin (2t)} \right)^2 \\
& = & \frac{{\sin ^2 (2t)}}{4} \\
\end{array}\)

 

[galactus]

Oh, I see. You had a sum the first post, not a product. A typo.

BTW, that's 'theta', not theda.

 

[Mrspi]

Yep thats right. The teacher comes up with really weird problems. We have to do this on our final

heres how the teacher typed it: SIN^4 THEDA COT^2 THEDA = 1/4 SIN ^2 (2) THEDA

The problem as you originally typed it had a "+" sign between sin<SUP>4</SUP> @ and cot<SUP>2</SUP> @.....that's quite a different problem.

sin<SUP>4</SUP> @ * cot<SUP>2</SUP> @ = (1/4) sin<SUP>2</SUP> (2@)

cot @ = cos @ / sin @, so we can re-write the left side:

sin<SUP>4</SUP> @ * (cos<SUP>2</SUP> @ / sin<SUP>2</SUP> @)

or,

sin<SUP>2</SUP> @ * cos<SUP>2</SUP> @

Multiply by 4/4:

(4 sin<SUP>2</SUP> @ cos<SUP>2</SUP> @) / 4

(2 sin @ cos @)<SUP>2</SUP> / 4

(sin 2@)<SUP>2</SUP> / 4

(1/4) * (sin 2@)<SUP>2</SUP>

And that's what is on the right-hand side.