Question about Computers & Internet

Your Question :

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

Hi,

a 6ya expert can help you resolve that issue over the phone in a minute or two.

best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.

the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).

click here to download the app (for users in the US for now) and get all the help you need.

goodluck!

Posted on Jan 02, 2017

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.

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

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.

Apr 01, 2014 | SoftMath Algebrator - Algebra Homework...

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

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

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).

Nov 07, 2010 | SoftMath Algebrator - Algebra Homework...

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

(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)]

(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'

But

- (29)'=0 derivative of a constant is zero
- (sin(2X))'=cos(2X)*(2X)'=2*cos(2X)
- (sin(X))'=cos(X)

(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)]

Jun 21, 2010 | Vivendi Excel@ Mathematics Study Skills...

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

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...

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)

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)

May 12, 2009 | ValuSoft Bible Collection (10281) for PC

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

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

Feb 02, 2009 | Super Tutor Trigonometry (ESDTRIG) for PC

THIS PROBELM IS TO DIFFICULT SO PLEASE SLOVE THIS PROBELM

Oct 07, 2008 | Computers & Internet

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

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

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.

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

175 people viewed this question

Usually answered in minutes!

×