(Less than) 21 Trig Questions from Amrita

Bob · 2014-05-19 04:50:57

Hint for Stefy:

Bob

Amrita · 2014-05-19 05:39:19

Can u plz help me out with a sum?
If 3sinθ +4cosθ =5,then 4sinθ -3cosθ =?

bobbym · 2014-05-19 05:46:55

Hi;

Amrita · 2014-05-19 05:54:56

Hi Bobbym,plz show me this sum too.

bobbym · 2014-05-19 05:56:24

Hi;

Sorry, one problem is all my little brain can do right now. I am working on a weird solution to the other one.

ShivamS · 2014-05-19 05:56:49

Bobbym, the sine of an imaginary number is always imaginary.

Bob · 2014-05-19 06:05:56

hi Amrita,

This will involve the compound angle formulas as well. I cannot find them on MIF but they are here:

http://www.mathsrevision.net/advanced-l … e-formulae

This problem is ready made for these.

If 3sinθ +4cosθ =5, (then 4sinθ -3cosθ =?)

divide by 5

3/5 sint + 4/5 cos t = 1

So let 3/5 = cos a and 4/5 = sin a

Then we have

cos a . sin t + sin a . cos t = 1

Using the compound angle formula:

sin(t+a) = 1

Now go to the expression you have to evaluate:

4 . sin t - 3 . cos t = 5 x ( 4/5 . sint - 3/5 cos t) = 5 x (sin a . sin t - cos a . cos t)

The second CAF will help here cos a . cos t - sin a . sin t = cos (a + t), so the expression becomes

5 x ( -cos(t + a) )

Now if sin(t + a) = 1 [90 degrees] then cos (t + a) = 0

So the expression is 5 x 0 = 0

Bob

Last edited by Bob (2014-05-19 06:09:53)

Amrita · 2014-05-19 06:14:55

Thank u vry mch,Bob. But I could'nt understand why cos(a+t) = -cos(a+t) need 2 b shown in the 3rd last step?

Bob · 2014-05-19 06:28:22

The formula is cos cos - sin sin but we have the reverse of that: sin sin - cos cos. So I had to insert a minus sign.

Bob

Amrita · 2014-05-19 06:30:40

I understood it.Thnx.Can I give u another sum plz?

ShivamS · 2014-05-19 06:31:35

Hi Amrita;

you don't have to ask to ask.

Amrita · 2014-05-19 06:41:18

O.k. Shivam.

If a=b cos(2π /3)=c cos(4π /3) , then find: ab+bc+ac.

Bob · 2014-05-19 09:48:19

hi Amrita,

First a preliminary formula:

Now let A+B = P and A-B = Q. => 2A = P + Q => A = (P+Q)/2 and also 2B = P - Q => B = (P-Q)/2

Substituting this into the above gives one of the factor formulas:

So what use is this? Well you can write the cosine bits of the starting equations like this:

So add these together and use the factor formula:

You can finish it off from here ??

Bob

Agnishom · 2014-05-20 06:12:13

bob bundy wrote:

hi Amrita,
This will involve the compound angle formulas as well. I cannot find them on MIF but they are here:
http://www.mathsrevision.net/advanced-l … e-formulae
This problem is ready made for these.
If 3sinθ +4cosθ =5, (then 4sinθ -3cosθ =?)
divide by 5
3/5 sint + 4/5 cos t = 1
So let 3/5 = cos a and 4/5 = sin a
Then we have
cos a . sin t + sin a . cos t = 1
Using the compound angle formula:
sin(t+a) = 1
Now go to the expression you have to evaluate:
4 . sin t - 3 . cos t = 5 x ( 4/5 . sint - 3/5 cos t) = 5 x (sin a . sin t - cos a . cos t)
The second CAF will help here cos a . cos t - sin a . sin t = cos (a + t), so the expression becomes
5 x ( -cos(t + a) )
Now if sin(t + a) = 1 [90 degrees] then cos (t + a) = 0
So the expression is 5 x 0 = 0
Bob

I've a cool looking method

Squaring (3sinθ +4cosθ), you get 9 sin^2 θ + 16 cos^2 θ + 24 sinθcosθ
Squaring (4sinθ -3cosθ), you get 16 sin^2 θ + 9 cos^2 θ - 24 sinθcosθ
Thus, (3sinθ +4cosθ)^2 + (4sinθ -3cosθ)^2 = 9 sin^2 θ + 16 cos^2 θ + 24 sinθcosθ + 16 sin^2 θ + 9 cos^2 θ - 24 sinθcosθ = 25 sin^2 θ + 25 cos^2 θ = 25
Or, (3sinθ +4cosθ)^2 + (4sinθ -3cosθ)^2 = 25
We are given that, (3sinθ +4cosθ) is 5. Plugging this into the above identity, we shall get 4sinθ -3cosθ = 0

Bob · 2014-05-20 06:40:05

That IS cool. (n)ice.

Bob

Amrita · 2014-05-27 21:13:46

Prove that (2√ 3+3)sinθ +(2√ 3)cosθ lies between -(2√ 3+√ 15) and (2√ 3+√ 15)

Agnishom · 2014-05-28 18:15:31

Bump!

Bob · 2014-05-28 20:03:33

Any expression, E, of the form asin(θ) + bcos(θ) can be written as a single sine expression, which will have easy to find max and min values. Do this

R = root(a^2 + b^2). tan (alpha) = b/a

Then

E = R[cos(alpha).sinθ + sin(alpha).cos(θ)] = R.sin(θ+alpha)

As sine goes from -1 to + 1, E will go from -R to + R.

Bob

Amrita · 2014-05-30 00:15:50

If tan(a-b) = 1 , sec(a+b) = 2/√ 3 ; find smallest (+ve) value of b.

Agnishom · 2014-05-30 02:41:09

Didn't Amin Babu tell us that we don't have the general solutions this time?

Agnishom · 2014-05-30 02:54:06

Solving those trigonometric equations, you get:

Solving these,

Apparently, to get the mimum +ve value of can be obtained by doing the following substitution in the second equation, but I cannot explain why :

The guy from bumpkinland has done that in the next post. Maybe you can ask him.

Last edited by Agnishom (2014-05-30 03:53:35)

bobbym · 2014-05-30 02:54:35

Agnishom · 2014-05-30 03:03:43

Are you sure? Is what I've done so far correct?

bobbym · 2014-05-30 03:06:07

If c[1] and c[2] are natural numbers as you say then your b will always be negative and you are asked for the smallest positive solution.

Agnishom · 2014-05-30 03:08:40

They are non-negative integers. I updated the post.

What do I do to minimise them keeping b +ve?

Math Is Fun Forum

#76 2014-05-19 04:50:57

Re: (Less than) 21 Trig Questions from Amrita

#77 2014-05-19 05:39:19

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#78 2014-05-19 05:46:55

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#79 2014-05-19 05:54:56

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