1. ## sine help

Hello all... just took a test in my pre calc class and got it back with a few wrong. We can correct them and then retake the test but im having trouble with corrections.

These are the ones that im having trouble with:

solve the equation -

2cos(πx + π/3) = - sqrt(2)

sin2x = sin6x

sin2x = -sqrt(2) * sin x

Im aware of the formulas and ive tried using them for these but am not getting any success.

for the first one i used arccosine after dividing each side by 2.. easy solve after that... got 5/12 and 11/12 but was marked wrong

for the second one i was not to sure but i said 0 was the answer.

for the last one i tried dividing by sin x and then using the formula sine2x = 2 sine x cos x... then turning the left side to 2sin x tan x. After dividing by 2 you can solve but still got the wrong answers...

Help would be a great thing and thanks in advance for anyone who has any suggestions

2. Originally Posted by rm1991
Hello all... just took a test in my pre calc class and got it back with a few wrong. We can correct them and then retake the test but im having trouble with corrections.

These are the ones that im having trouble with:

solve the equation -

2cos(πx + π/3) = - sqrt(2)

sin2x = sin6x

sin2x = -sqrt(2) * sin x

Im aware of the formulas and ive tried using them for these but am not getting any success.

for the first one i used arccosine after dividing each side by 2.. easy solve after that... got 5/12 and 11/12 but was marked wrong

for the second one i was not to sure but i said 0 was the answer.

for the last one i tried dividing by sin x and then using the formula sine2x = 2 sine x cos x... then turning the left side to 2sin x tan x. After dividing by 2 you can solve but still got the wrong answers...

Help would be a great thing and thanks in advance for anyone who has any suggestions
for the last one $\displaystyle 2sin(x)cos(x)=-\sqrt{2}sin(x)\Rightarrow{2sin(x)cos(x)+\sqrt{2}si n(x)=0}$$\displaystyle \Rightarrow{sin(x)(2cos(x)+\sqrt{2})=0}$ now use the zero product property

3. thanks i get it... didnt see it like that at first

4. Originally Posted by rm1991
Hello all... just took a test in my pre calc class and got it back with a few wrong. We can correct them and then retake the test but im having trouble with corrections.

These are the ones that im having trouble with:

solve the equation -

2cos(πx + π/3) = - sqrt(2)

sin2x = sin6x

sin2x = -sqrt(2) * sin x

Im aware of the formulas and ive tried using them for these but am not getting any success.

for the first one i used arccosine after dividing each side by 2.. easy solve after that... got 5/12 and 11/12 but was marked wrong

for the second one i was not to sure but i said 0 was the answer.

for the last one i tried dividing by sin x and then using the formula sine2x = 2 sine x cos x... then turning the left side to 2sin x tan x. After dividing by 2 you can solve but still got the wrong answers...

Help would be a great thing and thanks in advance for anyone who has any suggestions
For the second one I would use $\displaystyle sin(6x)=sin(2x+4x)$ then go from there..and 0 would be an answer

5. hmmm makes since! thanks a lot

oh btw, there is another one about a triangle that i thought i had but seem to be lost with

We are not allowed to use a calculator to make calculations with this and he gave us an approximate value of 0.6 for cos 53... We were supposed to find the area of a triangle with one side 35 another side 4 and angle between them 53.

I tried using the law of cosines and herons theorem but theres a sqrt after you use law of cosines to find the other side... so i would assume that this is not the way to do it (since were not supposed to use a calculator to find the answer)

6. Originally Posted by rm1991
hmmm makes since! thanks a lot

oh btw, there is another one about a triangle that i thought i had but seem to be lost with

We are not allowed to use a calculator to make calculations with this and he gave us an approximate value of 0.6 for cos 53... We were supposed to find the area of a triangle with one side 35 another side 4 and angle between them 53.

I tried using the law of cosines and herons theorem but theres a sqrt after you use law of cosines to find the other side... so i would assume that this is not the way to do it (since were not supposed to use a calculator to find the answer)
$\displaystyle C^2=16+35^2-2\cdot{4}\cdot{3}5\cdot{.6}$...group and factor and its not that difficult...Heron's Formula...with this ! I am assuming it was multiple choice...regardless the operative word in the directions is...approximately

7. wait but the law is $\displaystyle c^2 = a^2 + b^2 - 2abCosC$... so where did the minus 6 come from? and no.. not multiple choice sadly

8. Originally Posted by rm1991
wait but the law is $\displaystyle c^2 = a^2 + b^2 - 2abCosC$... so where did the minus 6 come from? and no.. not multiple choice sadly
Aha! Check again...my typo was fixed...I am goood ...haha just kidding

9. hmm... ok

same as i got earlier but dont know how my teacher expects us how to do this with out a calculator... the computations done with heron's theorem must be hellish!

10. Originally Posted by rm1991
hmm... ok

same as i got earlier but dont know how my teacher expects us how to do this with out a calculator... the computations done with heron's theorem must be hellish!
Yes...they scare me...

11. by the way, for $\displaystyle sin2x = sin 6x$ i get $\displaystyle sin2x = sin6x \rightarrow sin2x = sin(2x + 4x) \rightarrow sin2x = sin2x * cos4x + cos2x * sin4x$ and im stuck there

12. cos 53=0.6
sin 53=sqrt(1-0.6^2)
=0.8
area=0.5*ab sin 53
=0.5*35*4*0.8
=56