Question

Use trigonometric substitution to write the algebraic expression as a trigonometric function of θ, where 0 < θ <π/2.

√(100) - 4x^(2), x = 5cos(θ)

Answer 1

I really need an explanation someone plz help me!

Answer 2

hmm is \[\sqrt{100} - 4x^2 \]can be written as \[10 - 4(5\cos \Theta)^2\]from substituting x into the equation and simplifying \[\sqrt{100} = 10\]

Answer 3

to get \[10 - 100\cos^2\Theta\]

Answer 4

is the equation equal to anything?

Answer 5

yes. the options are: a. 100sin^2Θ b. 10sinΘ c. 20sinΘ d. 10+5cosΘ

Answer 6

hmm im not really sure what the question is asking

Answer 7

\[\cos^2 \Theta \] can also be written as \[(1+\cos2\Theta)/2\]but using that in the equation with \[10 - 100\cos^2 \Theta\]only gives\[50\cos \Theta - 40\]

Answer 8

unlesss there is some mistake that i made

Answer 9

\[(-40 - 50\cos2\Theta)*\]

Answer 10

forgot the negative

Answer 11

now if the question asked if \[\sqrt{100} - 4x^2 = 100\sin^2θ, x = 5\cos(θ)\] that would make more sense

Answer 12

are the options that you gave solutions or multiple questions?

Answer 13

cause then from a)
you get \[100\cos^2\theta + 100\sin^2\theta = \sqrt{100}\]

Answer 14

and then you can use the fact that \[\sin^2\theta = (1-\cos2\theta)/2 \]and\[\cos^2\theta = (1+\cos2\theta)/2 \] and solve for the expression in terms of theta

Answer 15

solutions. it's a multiple choice question

Answer 16

hmm the closest i can seem to get is\[ −40−50\cos2\theta\]or\[-90 + 100\sin^2\theta\]

Answer 17

is that square root suppose to be there? cause if it wasnt then you can get \[100\sin^2\theta\]which is one of the options

Answer 18

you get this from\[ \sin^2 \theta + \cos^2 \theta = 1\] through some algebra \[\cos^2 \theta = 1 - \sin^2 \theta\]and substituting for\[ \cos^2 \theta\]

Answer 19

adedokun so the equation kandae wrote was correct interpretation of what you wrote?

Answer 20

or i mean expression

Answer 21

\[\sqrt{100-4x^2}=\sqrt{100-4(5\cos \theta)^2}=\sqrt{100-4*25\cos^2 \theta}=\sqrt{100}\sqrt{1-\cos^2 \theta}=10\sqrt{\sin^2\theta}=10\sin \theta\]

Answer 22

=\[10\sin \theta\]

Answer 23

it is important that you tell us what the square root is over otherwise you will not get what you are seeking