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Please help! Trigonometry questions

A couple of questions i'm stuck on:

Solve: sin(x+π/6)+sin(x−π/6)=3/2sin(x+ \pi /6)+sin(x- \pi /6)=\sqrt3/2

•

I used the addition formulas

•

Got an equation with sin and cos

•

Ended up with wrong answer



Solve: cosx+cos2x+cos3x=0cosx+cos2x+cos3x=0

•

I used the addition formulas

•

changed sin squared x to (1-cos squared x)

•

got a cubic equation

•

ended up with wrong answer



help is very much appreciated!
Reply 1
Avatar for Hopple
Hopple
18
Can you show your working so far?
Original post by number23
A couple of questions i'm stuck on:

Solve: sin(x+π/6)+sin(x−π/6)=3/2sin(x+ \pi /6)+sin(x- \pi /6)=\sqrt3/2

•

I used the addition formulas

•

Got an equation with sin and cos

•

Ended up with wrong answer



You obtain 4 terms from using the addition formulas. Two cancel each other, leaving you with two identical terms. Hopefully that helps, if not, post what you did.

Original post by number23

Solve: cosx+cos2x+cos3x=0cosx+cos2x+cos3x=0

•

I used the addition formulas

•

changed sin squared x to (1-cos squared x)

•

got a cubic equation

•

ended up with wrong answer



help is very much appreciated!


Splitting up the cos3x, you get cos(2x + x). You then write this in terms of cosx and cos2x, then substitute in expressions for cos2x and sin2x. If you did this and got it wrong, post your working.
(edited 11 years ago)
Reply 3
Avatar for number23
number23
OP
15
Original post by Hopple
Can you show your working so far?



Original post by dantheman1261
.


For the first one, I ended up with (sqrt3)sinx+cosx=(sqrt3)/2

The second, cos^3x + 3cos^2x +1 =0

Not sure if these are correct :s



The answers in the answer booklet, nπ+(−1)nπ/6n\pi +(-1)^n\pi/6

and for the second, nπ/2+π/4,2nπ±2π/3 n\pi/2+\pi/4, 2n\pi \pm 2\pi/3
(edited 11 years ago)
Original post by number23
For the first one, I ended up with (sqrt3)sinx+cosx=(sqrt3)/2

The second, cos^3x + 3cos^2x +1 =0



Could you write out the initial expansion for each? Something has gone wrong in both cases.
Reply 5
Avatar for Hopple
Hopple
18
Original post by number23
For the first one, I ended up with (sqrt3)sinx+cosx=(sqrt3)/2
You shouldn't get a cos(x) term at all, the two of them cancel each other out. The sin(x) term I agree with though.

The second, cos^3x + 3cos^2x +1 =0

I think you're missing quite a few terms here, maybe some algebraic slip with the mass of terms?
Reply 6
Avatar for number23
number23
OP
15
Original post by Hopple



Original post by dantheman1261


Thanks, i've managed to solve the first one.

For the second, do you know what cubic equation with cos i need to end up with? thanks
Reply 7
Avatar for Hopple
Hopple
18
Original post by number23
Thanks, i've managed to solve the first one.

For the second, do you know what cubic equation with cos i need to end up with? thanks


You can probably work it out from the answer given :wink: Just do it carefully, you know how to expand brackets, and you know the angle identities you need etc.

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