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### Section 7.3 : Trig Substitutions

6. Use a trig substitution to eliminate the root in $$\sqrt {1 - 4z - 2{z^2}}$$.

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Hint : This doesn’t look much like a term that can use a trig substitution. So, the first step should probably be to some algebraic manipulation on the quantity under the root to make it look more like a problem that can use a trig substitution.
Start Solution

We know that in order to do a trig substitution we really need a sum or difference of a term with a variable squared and a number. This clearly does not fit into that form. However, that doesn’t mean that we can’t do some algebraic manipulation on the quantity under the root to get into a form that we can do a trig substitution on.

Because the quantity under the root is a quadratic polynomial we know that we can complete the square on it to turn it into something like what we need for a trig substitution.

Here is the completing the square work.

\begin{align*}1 - 4z - 2{z^2} & = - 2\left( {{z^2} + 2z - \frac{1}{2}} \right)\hspace{0.75in}{\left[ {\frac{1}{2}\left( 2 \right)} \right]^2} = {\left[ 1 \right]^2} = 1\\ & = - 2\left( {{z^2} + 2z + 1 - 1 - \frac{1}{2}} \right)\\ & = - 2\left[ {{{\left( {z + 1} \right)}^2} - \frac{3}{2}} \right]\\ & = 3 - 2{\left( {z + 1} \right)^2}\end{align*}

So, after completing the square the term can be written as,

$\sqrt {1 - 4z - 2{z^2}} = \sqrt {3 - 2{{\left( {z + 1} \right)}^2}}$
Hint : At this point the problem works in the same manner as the previous problems in this section.
Show Step 2

So, in this case we see that we have a difference of a number and a squared term with a variable in it. This suggests that sine is the correct trig function to use for the substation.

Now, to get the coefficient on the trig function notice that we need to turn the 2 (i.e. the coefficient of the squared term) into a 3 once we’ve done the substitution. With that in mind it looks like the substitution should be,

$z + 1 = \frac{{\sqrt 3 }}{{\sqrt 2 }}\sin \left( \theta \right)$

Now, all we have to do is actually perform the substitution and eliminate the root.

Show Step 3
\begin{align*}\sqrt {1 - 4z - 2{z^2}} & = \sqrt {3 - 2{{\left( {z + 1} \right)}^2}} = \sqrt {3 - 2{{\left( {\frac{{\sqrt 3 }}{{\sqrt 2 }}\sin \left( \theta \right)} \right)}^2}} \\ & = \sqrt {3 - 3{{\sin }^2}\left( \theta \right)} = \sqrt 3 \sqrt {{{\cos }^2}\left( \theta \right)} = \require{bbox} \bbox[2pt,border:1px solid black]{{\sqrt 3 \left| {\cos \left( \theta \right)} \right|}}\end{align*}

Note that because we don’t know the values of $$\theta$$ we can’t determine if the cosine is positive or negative and so cannot get rid of the absolute value bars here.