Question

can someone please help me with trigonometric substitutions?

Answer 1

it's much easier if you just post the specifics

Answer 2

I do that and no one answers me

Answer 3

sometimes we may not know or are just caught up on something at the moment, we can always check back on material posted

Answer 4

Use trigonometric substitution to write 3= radical (36-x^2) as a trigonometric function of theta, where -pi/2<theta<pi/2 given x=6 sin theta. Use the function to find the value of sin theta.

Answer 5

hmmm not sure I follow that... can you post a quick screenshot of the material?

Answer 6

do you have any idea how to solve that?

Answer 7

I'm incredibly lost.

Answer 8

\(\bf 3 = \sqrt{(36-x^2)}\qquad \qquad x = 6sin(\theta)\\ \quad \\
3 = \sqrt{(36-x^2)}\implies 3 = \sqrt{(36-(6sin(\theta))^2)}\\ \quad \\
\textit{squaring both sides}\\ \quad \\
(3)^2 = \left(\sqrt{(36-(6sin(\theta))^2)}\right)^2\implies (3)^2 = \sqrt{[36-(6sin(\theta))^2]^2}\\ \quad \\
9=36-(6sin(\theta))^2\implies (6sin(\theta))^2=36-9\implies (6sin(\theta))^2=27\\ \quad \\
\textit{now taking square root to both sides}\\ \quad \\
\sqrt{(6sin(\theta))^2}=\sqrt{27}\implies 6sin(\theta)=\sqrt{27}\\ \quad \\
sin(\theta)=\cfrac{\sqrt{27}}{6}\)

Answer 9

\(\bf \color{blue}{27 = 3\times 3\times 3\implies 3^2\times 3}\qquad thus\\ \quad \\
sin(\theta)=\cfrac{\sqrt{27}}{6}\implies sin(\theta)=\cfrac{\sqrt{3^2\times 3}}{6}\implies sin(\theta)=\cfrac{\cancel{3}\sqrt{ 3}}{\cancel{6}}\\ \quad \\
\implies sin(\theta)=\cfrac{\sqrt{ 3}}{2}\)

Answer 10

thank you so much! I actually understood that!

Answer 11

yw