Question

Use the squared identities to simplify 2sin^2sin^2x

Answer 1

@directrix

Answer 2

u mean
sin^2(a) + cos^2(a) = 1 ?

Answer 3

>> squared identities also known as Pythagorean Identities

Answer 4

@DanJS I don't know honestly that's how the problem is written and my answer choices are

A.3+4cos(2x)+cos(4x)/4
B.3-4cos(2x)+cos(4x)/4
C.3-4cos(2x)-cos(4x)/4
D.3+4cos(2x)-cos(4x)/4

Answer 5

\[2\sin^{2}(\sin^{2}(x)) \]

Answer 6

that what you say?

Answer 7

I'm attached the "Squared Indentities" here for you to review.
And, we'll probably need them here.

But, the way you posted this problem, it is hard to tell what is an angle and what we are finding. Use grouping symbols often for clarity.

Answer 8

The problem is \[2\sin^2x \sin^2x\]

Answer 9

Okay, now we can get to work.

Answer 10

oh, that makes more sense

Answer 11

Sorry about that guys

Answer 12

No worries.

Answer 13

so it is just \[2 \sin^{4}(x) \]
not a compound sine

Answer 14

looking at the answers, it seems you need to apply the half angle formulas for sin^2(x)

Answer 15

\[\sin ^{2}x = \frac{ 1-\cos(2x) }{ 2 }\]

Answer 16

how do I apply the half angle formula? @DanJS

Answer 17

Do you have any idea how to do this? @Directrix I'm stuck :/

Answer 18

Did you try taking 2* (sinx)^2 * (sinx)^2 and substituting( (1 - cos(2x) )/2 for each of the
(sinx)^2 terms and then multiplying those?

Answer 19

@Directrix Did you get my message?

Answer 20

That solution looks right and yields the correct answer. Impressive formatting.

Answer 21

@Monster11223 ^^

Answer 22

@Directrix Thanks and it took me a while to do it lol well thats it for now thanks a lot :D

Answer 23

I'm pleased with your work. See ya later on the OS. Wishing you the best on your exams.

Trig always take a lot of time!