Verify the Identity

[math-a-phobic]

Please help! I tried to verify the identity for these four problems but got stuck Please check to see what I did wrong.



Problem 1:

Problem 2 and Problem 3:

Problem 4:

Thanks for the help!

 

[galactus]

#1:

Factor out \(\displaystyle \L\\sin^{\frac{1}{2}}(x)cos(x)\):

\(\displaystyle \L\\sin^{\frac{1}{2}}(x)cox(x)(1-sin^{2}(x))\)

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

\(\displaystyle \L\\sin^{\frac{1}{2}}(x)cos(x)(cos^{2}(x))=\)

\(\displaystyle \L\\sin^{\frac{1}{2}}(x)cos^{3}(x)\)

 

[soroban]

Hello, math-a-phobic!

I like your substitutions . . .
\(\displaystyle \;\;\) Let \(\displaystyle x\,=\,\sin\theta,\;\;y\,=\,\cos\theta\)
and remember that: \(\displaystyle \,x^2\,+\,y^2\:=\:1\)

\(\displaystyle \left(\sin\theta\right)^{\frac{1}{2}}\cos\theta \,-\,\left(\sin\theta\right)^{\frac{5}{2}}\cos\theta \;= \;\cos^3\theta\sqrt{\sin\theta\)

We have: \(\displaystyle \,x^{\frac{1}{2}}y\,-\,x^{\frac{5}{2}}y\)

Factor: \(\displaystyle \,x^{\frac{1}{2}}y\,\left(\underbrace{1\,-\,x^2})\)
Then: \(\displaystyle \;\;\;x^{\frac{1}{2}}y\,\cdot\, y^2\;= \;y^3\sqrt{x}\;\;\Rightarrow\;\;\cos^3\theta\,\sqrt{\sin\theta}\)

\(\displaystyle \cos\theta \,- \,\L\frac{\cos\theta}{1\,-\,\tan\theta} \;= \;\frac{\sin\theta\cos\theta}{\sin\theta\,-\,\cos\theta}\)

We have: \(\displaystyle \,y \,-\,\L\frac{y}{1\,-\,\frac{x}{y}}\)

Factor: \(\displaystyle \L\,y\left(1\,-\,\frac{1}{1\,-\,\frac{x}{y}}\right)\)

Multiply top and bottom of the fraction by \(\displaystyle y:\L\;y\left(1\,-\,\frac{y}{y\,-\,x}\right)\)

Combine: \(\displaystyle \L\,y\left(\frac{y\,-\,x\,-\,y}{y\,-\,x}\right)\;=\;\frac{-xy}{y\,-\,x}\;=\;\frac{xy}{x\,-\,y}\;\;\Rightarrow\;\;\frac{\sin\theta\cos\theta}{\sin\theta\,-\,\cos\theta}{\)

\(\displaystyle 2\,+\,\cos^2\theta\,-\,3\cos^4\theta\;=\;\sin^2\theta(2\,+\,3\cos^2\theta)\)

On the right, we have: \(\displaystyle \,x^2(2\,+\,3y^2)\)
\(\displaystyle \;\;\)Then we have: \(\displaystyle \;\;\,(\overbrace{1\,-\,y^2})(2\,+\,3y^2)\)

Multiply: \(\displaystyle \,2\,+\,y^2\,-\,3y^4\;\;\Rightarrow\;\;2\,+\,\cos^2\theta\,-\,3\cos^4\theta\)