Can you figure out this trigonometry problem?

Have you written it properly? If you do:

8 x sin(10°) * sin(30°) * sin(50°) * sin(70°) as a straight multiplication problem, then the answer is .5 (actually, .4999999999999…)

See here

Yes, but how can you prove it?

Should there be an end parenthesis after 30?

And what does the question want you to do? There’s no variable to solve for. Do you just want a proof that that product equals ½?

Yes there should be a parenthesis after 30. And the problem asked for a proof that the product is ½.

I think your identity sin2x=2sinx cos x is the key…

I would convert the sines into cosines…

Cos20.cos40.cos60.cos80 =1/16

and substitute in using your formula so cos20 becomes 1/(2sin20).sin40 and cos60=½

The sin40 combines with the cos40 to give sin80
The sin 80 combines with the cos80 to give sin120
Sin120 = sin20 and so cancels out with the 1/sin20

And you have eliminated all the sines and cosines.

@flutherother Brilliant, GA.

I agree.

Can a high school student be expected to solve this?

Once you see the answer, it’s not really hard, doesn’t involve any very complex math, you just need a bit of creativity. I don’t think just any high school student would come up with it, but some could.

There was another problem that went along with this one:
8cos(10)cos(30)cos(50)cos(70)=3/2

I tried converting it to
8sin(80)sin(60)sin(40)sin(20), but @flutherother‘s trick does not work for sin

The key to this one is CosAcosB = ½(cos(A+B)+cos(A-B))

(cos70cos10)(cos50cos30)
=½(cos80+cos60)½(cos80+cos20)
=¼(cos2(80)+cos80cos20+cos80cos60+cos60cos20)
=1/8(1.5+cos160+cos100+cos140+cos20+cos80+cos40)
=3/16+1/8(cos160+cos20)+1/8(cos140+cos40)+1/8(cos100+cos80)
As cosA+cosB=2cos1/2(A+B)cos1/2(A-B) the terms in brackets are 0
=3/16

You have done it again! This time I really have to question how many high schoolers could figure this one out.

If high schoolers had been practising examples that use these identities they would be on the right track from the start. Or there may be more straightforward solutions.

I remember when I was a high school kid our trigonometry teacher taught us two useful identities:

4 sin(a) sin(60-a) sin(60+a) = sin(3a)
and
4 cos(a) cos(60-a) cos(60+a) = cos(3a)

The identities aren’t hard to prove. Just use product identities a few times:

4 sin(a) sin(60-a) sin(60+a) = 2 sin(a) ( cos(2a) – cos(120)) = sin(3a) – sin(a) – 2 sin(a) cos(120) = sin(3a) – sin(a)+sin(a) = sin(3a)

Similarly you can get the second identity.

Now if you replace a by 10 you get the two identity that you are looking for.

I had never realized this but now that I think about the above two identities it sounds like these two identities can be generalized for any n, so that we have product of n trig functions on the left and a trig function of na on the right. (in the above identity we have n=3) I will post the identity and a neat proof of that if there is any interest.

I would be definitely be interested.

I posted the identities and proof of one of them here. The other one can be proved similarly.

Edit: oops. The second identity doesn’t seem to be correct for all n’s. If you replace a by a+pi/2 in the first identity on the left you get product of cosines but on the right you get sin(na+npi/2) which is not always equal to cos(na).

Thanks. I will be looking at it today as soon as I can shake off the weekend doldrums.