Question

If x = 7 sin θ, write the expression θ/2 − sin 2θ/4 in terms of just x. (Simplify the double angle using a double-angle identity.)

Answer 1

you can simplify further
also the minus shouldn't turn into plus

Answer 2

\[\text{ recall } \frac{x}{7}=\sin(\theta) \text{ use a right triangle }\]
\[\text{ this will help you find } \cos(\arcsin(\frac{x}{7}))\]

Answer 3

okok

Answer 4

ok so now I have 1/2(sin^-1(x/7))(sqrt49-x^2 /7)

Answer 5

something seems to be missing

Answer 6

because you had \[\frac{1}{2}[ \arcsin(\frac{x}{7})-\frac{x}{7} \cos(\arcsin(\frac{x}{7})] \\ \frac{1}{2}[\arcsin(\frac{x}{7})-\frac{x}{7} \frac{\sqrt{49-x^2}}{7}]\]

Answer 7

of course you can multiply that 7 and the other 7

Answer 8

yeah so would it be \[\frac{ 1 }{ 2 } \arcsin(\frac{ x }{ 7}) - \frac{ x \sqrt{49-x ^{2}} }{ 49 }\]

Answer 9

well if you write it like that you need to distribute the 1/2 to both terms

Answer 10

well parentesis after 1/2

Answer 11

not just the first

Answer 12

the parenthesis didnt come out for some reason

Answer 13

ah okay

Answer 14

so would that be right with the parethesis?

Answer 15

or should I multiply the 2 to both denominators so that it is over 14 and over 98

Answer 16

I don't see how you would get the 14

Answer 17

If you distribute...
\[\frac{1}{2}[ \arcsin(\frac{x}{7})-\frac{x}{7} \cos(\arcsin(\frac{x}{7})] \\ \frac{1}{2}[\arcsin(\frac{x}{7})-\frac{x}{7} \frac{\sqrt{49-x^2}}{7}] \\ \frac{1}{2}\arcsin(\frac{x}{7})-\frac{x \sqrt{49-x^2}}{98}\]

Answer 18

so is that right?

Answer 19

nope it was wrong

Answer 20

it is not wrong

Answer 21

\[\sin(\theta)=\frac{x}{7}\]
\[\theta=\arcsin(\frac{x}{7}) \\ \frac{\theta}{2}-\frac{\sin(2 \theta)}{4} =\frac{1}{2} \arcsin(\frac{x}{7})-\frac{2 \sin(\theta)\cos(\theta)}{4} \\ =\frac{1}{2}\arcsin(\frac{x}{7})-\frac{1}{2}\sin(\theta)\cos(\theta) \\ =\frac{1}{2} \arcsin(\frac{x}{7})-\frac{1}{2} \frac{x}{7} \frac{\sqrt{49-x^2}}{7} \\ = \frac{1}{2} \arcsin(\frac{x}{7})-\frac{x \sqrt{49-x^2}}{98}\]