Question

Hey! I was hoping for someone to check my answer?

I need to fill in the missing blanks to these two problems:



1. _ _ _(pi/3)sin(pi/6)=1/2(sin(pi/2)-sin(pi/6))

For the missing part, I thought the answer would just be sin??

2. sin(pi/4)sin(pi/6)=1/2(_ _ _(pi/12)-cos(5pi/12))

I was thinking this one's missing part is: cos??

Help please!

Answer 1

@ganeshie8 Help por favor? :)

Answer 2

You're using these formulas :
http://prntscr.com/3rd6ss

right ?

Answer 3

Attempting too, yes... For some reason. Filling in the blanks is really confusing for me.. Too check myself.. I've been using this calculator.. Which probably isn't very helpful in learning how to solve...

http://symbolab.com/solver/trigonometric-identity-calculator

Answer 4

I see... it will become easy if we break it down into 1-2 steps :

_ _ _(pi/3)sin(pi/6)=1/2(sin(pi/2)-sin(pi/6))

is same as :

_ _ _(pi/3)sin(pi/6)=1/2(sin(pi/3 + pi/6)-sin(pi/3 - pi/6))

Answer 5

Oh! So I would solve the equation on the right to get the missing part on the left?

Answer 6

yes :
sin(A+B) = sinAcosB + cosAsinB
sin(A-B) = sinAcosB - cosAsinB
-----------------------------------
sin(A+B) - sin(A-B) = 2cosAsinB

Answer 7

so you should fill the first blank by `cos`

Answer 8

Oh!! That makes sense. so it'd look like: cos(pi/3)sin(pi/6)=1/2(sin(pi/2)-sin(pi/6))?

Answer 9

http://www.wolframalpha.com/input/?i=cos%28pi%2F3%29sin%28pi%2F6%29%3D1%2F2%28sin%28pi%2F2%29-sin%28pi%2F6%29%29

Answer 10

Yep !

Answer 11

Yay! for cos being for the first!

Answer 12

second one is correct ! it has to be `cos`

Answer 13

http://prntscr.com/3rd6ss

look up second identity ^

Answer 14

Whoops Idk why I sent that again

Answer 15

I was gonna ask.. Could you help me with using squared identities?

Answer 16

to be honest - i always get confuse with these product formulas pretty badly, and always derive them from the sum formulas when needed. We only need to remember four formulas :

```
sin(A+B) = sinAcosB + cosAsinB
sin(A-B) = sinAcosB - cosAsinB
cos(A+B) = cosAcosB - sinAsinB
cos(A-B) = cosAcosB + sinAsinB
```

Answer 17

yeah sure, ask...

Answer 18

Thank you for those formula's.. I saved them..

Okay so I'm supposed to use square identities to simplify sin^2sin^2x...

I'm pretty lost.

Answer 19

I have answer choices if that'd help a bit.

Answer 20

Is the expression just that ? \(\large \sin^2 (\sin^2x)\) ?

yeah provide the answer choices...

Answer 21

No, it just looks like: sin^2x sin^2x So it's just repeated.. It'll take me second to type them out! Don't run away!

Answer 22

I'll be here - may be take a snap and attach the file if possible

Answer 23

K! Here it is!

Answer 24

we use below two formulas :
\(\large \sin^2 x = \dfrac{1-\cos (2x)}{2}\)

\(\large \cos^2 x = \dfrac{1+\cos (2x)}{2}\)

Answer 25

\(\large \sin^2x\times \sin^2x = \left(\sin^2x\right)^2 = \left(\dfrac{1-\cos(2x)}{2}\right)^2 \)

Answer 26

\(\large = \dfrac{1}{4}\left(1-\cos(2x)\right)^2 \)

Answer 27

use (a-b)^2 formula and expand...

Answer 28

\(\large = \dfrac{1}{4}\left(1-2\cos(2x) + \cos^2(2x)\right) \)

Answer 29

use the second formula for cos^2 term

Answer 30

\(\large = \dfrac{1}{4}\left(1-2\cos(2x) + \dfrac{1+\cos(4x)}{2}\right) \)

Answer 31

simplify

Answer 32

Oh!! So I simplify that and it'll give me one of the answer choices? Seems simple enough! mmkay I'm gonna attempt to simplify. Just a sec

Answer 33

just pull out the 1/2 :

\(\large = \dfrac{1}{8}\left(2-4\cos(2x) + 1+\cos(4x) \right) \)

Answer 34

Oh, gotcha. So would that be sin^2x sin^2x simplified??? The reason I ask is because the answers I have to choose from are different.

Answer 35

add 2+1 ^

Answer 36

\(\large = \dfrac{1}{8}\left(3-4\cos(2x) +\cos(4x) \right) \)

Answer 37

Ahh, so the process keeps going? Geez no wonder I am terrible at math. So, the answer would be B! (3-4cos(2x)+cos(4x))/8 !!!

Answer 38

lol you're very good at math.. yes B is the answer - lets verify it with wolfram :
http://www.wolframalpha.com/input/?i=simplify+sin%5E2x*sin%5E2x

Answer 39

Yay for wolfram!! You're so helpful, thank you for putting up with my slowness!

Answer 40

np, you're welcome :)