Anyone good at Trigonometry?

Question

I have 2 problems I'm a little stumped on. The first one is:

sin^3 θ cos θ - sin θ cos^3 θ = -1/4

I think I should probably use the sum and difference equations for this one, but I'm not sure.

My other problem is:

sec 4θ + 2 sin 4θ = 0

This is what I've come up with so far, but I think it's wrong:

(1/cos 4θ) + 2 sin 4θ = 0
2 sin 4θ/ cos 4θ = 0

so, 2 tan 4θ = 0

(that's what I have so far)

If anyone can help with either of these problems, I would be very apperciative!

Thanks for your help!

Answer 1

Hi, on your first one,
sin^3 θ cos θ - sin θ cos^3 θ = -1/4

Factor out sin θ cos θ (sin^2 θ - cos^2 θ) = -1/4
You will notice that if the part in parentheses were reversed it would be an identity for cos 2θ so multiply both sides by -1 to have this happen:

sinθ cosθ (cos^2θ - sin^2θ) = 1/4

You will also notice if there were a 2 in front you'd have an identity for sin 2θ so multiply both sides by 2 to get

2 sinθcosθ (cos^2θ - sin^2θ) = 1/2

replace with the identities: sin 2θ cos2θ = 1/2

Now, if this had a 2 in front it would be an identiti for sin 2(2θ) so mult. both sides by 2 to get
2 sin 2θ cos 2θ = 1

sin 2(2θ) = 1

sin 4θ = 1

You know that sin pi/2 = 1 so 4θ = pi/2 and θ = pi/8. You can figure the additional answers as θ increases.


If I get the second I'll edit this unless someone else already got it

Here's what I think:
sec 4θ + 2 sin 4θ = 0

1/cos4θ + 2 sin 4θ = 0

get common denominators

1/cos 4θ + 2 sin4θ cos4θ/cos 4θ

you will see that the top of the second fraction is an identity for sin 2(4θ) so replace it

1/cos4θ + sin 2(4θ)/cos4θ = 0

This is 1 + sin 8θ all over cos 4θ = 0

If a fraction = 0, its numerator = 0 so 1 + sin 8θ = 0

sin 8θ = -1

sin of 3pi/2 = -1 so 8θ = 3pi/2 and θ = 3pi/16

again there are infinitely many more answers unless you were given a specific domain for θ

Hope that helps

Answer 2

lol...for some reason, i have a feeling we are in the same class!!!...but anyways...

the first one!!!
sin^3θcosθ - sinθcos^3θ = -1/4
sinθcosθ (sin^2θ -cos^2θ) = -1/4

=====TAKE THE NEGATIVE FROM THE INSIDE OUT!! =====MAKES IT EASIER

- sinθcosθ (sin^2θ - cos^2θ) = -1/4
- sinθcosθ (cos2θ) = -1/4

========MULTIPLY BOTH SIDES BY 2

-2sinθcosθ (cos2θ)= -1/2

=========2SINθCOSθ = SIN2θ SO......

-sin2θ*cos2θ= -1/2

==========MULTIPLY BOTH SIDES BY 2 ONCE AGAIN

-2sin2θ*cos2θ = -2/2

==========2SIN2θ*COS2θ = SIN 4θ SO........

-sin4θ= -1

sin4θ=1
4θ = pi/2
θ = pi/2 divided by 4
θ = pi/8


THIS IS WHAT I GOT FOR YOUR FIRST QUESTION...NOT TOTALLY SURE IF ITS RIGHT.BUT I TOOK A LONG TIME DOING IT , AND IT SEEMS TO BE RIGHT


for your second problem....

sec4θ +2sin4θ = 0
sec4θ= -2sin4θ

===================SEC4θ = 1/COS(4θ)

1/cos(4θ) = -2sin4θ
1= -2sin(4θ)*cos(4θ)
-1/2=sin(4θ)*cos(4θ)
-1/2=1/2[sin(4θ+4θ) + sin(4θ +4θ)]
-1/2 = 1/2[sin(8θ)+sin(0)]
-1/2= 1/2 sin(8θ)

==============multiply both sides by 2

-1 = sin(8θ)
sin(8θ) = 1
8θ = -pi/2
θ = -pi/2 divided by 8
θ = -pi/16

thatsthe answer i got for this one also...i also cannot guarantee that the answer its right..but VERY likely...

Answer 3

good luck