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VECTORS WHAT IS THE REASON FOR TO DEFINE a.b=|a||b| cos(a,b)

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Your Question :
VECTORS - 37.gif
My Answer ::
In scalar multiplication we are considering only logical values and doing multiplication but here we are taking care of the direction of the force magnitude of the force applied to the body for the perfect accurate value which helps in making of the real world application.

here scalar values will not work for making any real thing, So in your question cos (a,b) suggests the resultant direction of the resultant force and |a||b| are the magnitude.

For more details on vector multiplication :: http://mathworld.wolfram.com/VectorMultiplication.html

Posted on Sep 20, 2008

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Cos2x + 3 = 5cosx


Use the identity cos(2x)=2(cos(x))^2-1
cos(2x)+3=5cos(x) becomes 2(cos(x))^2-1+3=5cos(x)
Arrange a bit: 2(cos(x))^2-5cos(x)+2=0
Get rid of the 2-factor
(cos(x))^2-(5/2) cos(x)+1=0
This is a quadratic equation for the unknown U=cos(x)
U^2-(5/2)U+1=0
Solve it by factoring or with the quadratic equation formula. The solutions are U=2 or U=1/2.
Since U=cos(x), the root U=cox(x)=2 must be rejected.
What is left is cos(x)=(1/2). The solutions are x=60 or x=-60 plus or minus 360 degrees.

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1 Answer

Find the trig function given its period cos 5 pi


cos(5PI)=cos(4PI+PI)=cos(PI)=-1
sin(19PI/6)=sin(18PI/6+ PI/6)=sin(3PI +Pi/6)=sin(2PI+PI+PI/6)=sin(PI+PI/6)=sin(-PI/6)=-sin(PI/6)=-1/2

Dec 12, 2011 | Super Tutor Trigonometry (ESDTRIG) for PC

1 Answer

I need to rewrite Y=5(sqrt2)sin(x)-5(sqrt2)cos(x) as Y=Asin(Bx-c)


Use the fact that cos(pi/4)=sin(pi/4)= 1/square root(2). Trigonometric identity cos(a+b)=cos(a)cos(b)-sin(a)sin(b).

1_22_2012_4_04_59_am.jpg

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1 Answer

Differentiate each of the following w.r.t.x; 29.sin2xsinx


Use the rule for differentiating products of functions: ()' signifies derivative
(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'
But
  1. (29)'=0 derivative of a constant is zero
  2. (sin(2X))'=cos(2X)*(2X)'=2*cos(2X)
  3. (sin(X))'=cos(X)
Result is
(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)

You could also have cast your formula in the form
sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]
then calculated the derivative of
29/2*[cos(X)-cos(3X)]
which is
29/2*[-si(X) +3*sin(3X)]

The challenge for you is to prove that the two forms are equivalent
29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]

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1 Answer

Cos2x + 3 = 5cosx


the solution you got is correct it is

cos X =2 and cos X =1/2


but we know maximum value of cosX is 1.so we discard the solution cos X =2


so only solution is cos X=1/2

and X = 60 degrees

hope you r satisfied.please rate the solution high.thank you

Apr 17, 2010 | SoftMath Algebrator - Algebra Homework...

1 Answer

(1+cotx-cosecx)(1+tanx+secx)=2


I shall attempt :D
1) cosec A + cot A = 3
we know that (cot A)^2 + 1 = (cosec A)^2
Hence, (cosec A)^2 - (cot A)^2 = 1
thus, (cosec A + cot A) (cosec A - cot A) = 1
3 (cosec A - cot A) = 1
(cosec A - cot A) = 1/3

(cosec A - cot A) = 1/3
(cosec A + cot A) = 3
Summing them, 2 cosec A = 3 1/3
cosec A = 6 2/3 = 5/3
sin A = 0.15
Thus, cos A = sqrt (1 - (sin A)^2) = 0.989


2) Prove that (1+tan x - sec x)(1 + cot x + cosec x) =2
expand
LHS= 1 + cot x + cosec x + tan x + 1 + tan x cosec x - sec x - sec x cot x - sec x cosec x
We can calculate that
tan x cosec x = sec x (since tan x = sin x / cos x)
sec x cot x = cosec x
so the above is
LHS = 1 + cot x + cosec x + tan x + 1 + sec x - sec x - cosec x - sec x cosec x
LHS = 2 + cot x + tan x - sec x cosec x
LHS = 2 + cos x / sin x + sin x / cos x - 1 / (sin x cos x)
LHS = 2 + [{cos x}^2 + {sin x}^2 - 1] / (sin x cos x)
LHS = 2 (proved)

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1 Answer

Help


sec^4X- sec^2X = 1/cot^4X + 1/cot^2X
RHS
1/cot^4X + 1/cot^2X
=1/(Cos^4X/Sin^4X) + 1/(Cos^2X/Sin^2X)
=Sin^4X/Cos^4X + Sin^2X/Cos^2X
=Sin^4X/Cos^4X + Cos^2X.Sin^2X/Cos^4X
=Sin^2X/Cos^4(Sin^2X + Cos^2X)
=Sin^2X/Cos^4X
=(1-Cos^2X)/Cos^4X
=1/Cos^4X - Cos^2X/Cos^4X
=1/Cos^4X - 1/Cos^2X
=Sec^4X - Sec^2X
=LHS

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1 Answer

Trignometery: prove that .....


THIS PROBELM IS TO DIFFICULT SO PLEASE SLOVE THIS PROBELM

Oct 07, 2008 | Computers & Internet

1 Answer

Solve it plz......


cos x + root 3 sin x =root 2
cos x + ö3 * Sin x = ö2
squaring both the side
(cos x + ö3 * Sin x)2 = (ö2)2
Cos2 x + 3 * Sin2 x = 2
Cos2 x + Sin2 x + 2 * Sin2 x = 2
1 + 2 * Sin2 x= 2
2 * Sin2 x = 2-1
2 * Sin2 x = 1
Sin2 x = ½
Sin x = ö½
Sin x = 1/V2= Sin 45
X = 450

Aug 28, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

4 Answers

Trig Identities


Change csc to 1/sin. Find a common denominator and add the two left terms.
1/sin - sin = (1 -sin^2)/sin. Rewrite formula
(1 - sin^2)/sin = cos^2/sin Divide out the /sin.
1 - sin^2 = cos^2 Rearange.
1 = cos^2 + sin^2 Yes, that's true. It's like the Pythagorean formula.

May 22, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

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