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Trignometery: prove that .....

Square root of 1 - cos x / 1 + cos x = cosec x - cot x = 1 + sin x - cos x / 1 + sin x + cos x

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THIS PROBELM IS TO DIFFICULT SO PLEASE SLOVE THIS PROBELM

Posted on Oct 22, 2008

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Csc(x)cot(x)/sec(x)


csc(x)=1/sin(x)
sec(x)=1/cos(x)
csc(x)/sec(x)=(1/sin(x))*(cos(x)=cot(x)
csc(x)*cot(x)/sec(x)=(cot(x))^2=(tan(x))^(-2)

Jul 12, 2014 | Super Tutor Trigonometry (ESDTRIG) for PC

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Cot(x) = -0.6


Let's start with a little background.

The cot(x) is also known as the cotangent(x) and it equals 1/tan(x) which equals cos(x)/sin(x). I'm showing these formulas because your calculator may not have a cot button but it will probably have buttons for tan, cos, and sin.

Your calculator may also have buttons for tan-1, cos-1 and sin-1. These are the inverse functions for tan, cos, and sin. If you enter a number and then push the tan-1 button, the result is the angle whose tangent is the entered number. For example, it you enter 1 and push the tan-1 button the answer will be 45 deg because tan (45 deg) = 1.

Now let's look at the problem, cot(x) = -0.6.
The first thing we need to know is do you want the answer in degrees or radians? Your calculator will have both modes. The default mode when you first turn it on is probably degrees. If this problem is in radians you will need to change the mode of your calculator over to radians before we start.

If cot(x) = -0.6, then tan(x) = 1/-0.6 from the formula I showed in the background section.

This means tan(x) = -1.6666666...

Now we just enter -1.66666667 and hit the tan-1 button to get the answer.

If we're operating in radians the answer is -1.0307 radians.
If we're operating in degrees the answer is -59.036 deg.

I hope this helps you out.

Dec 06, 2011 | SoftMath Algebrator - Algebra Homework...

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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

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

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)]

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

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(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)

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

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

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

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

1 Answer

Trigonometry proofs


tan y/sin y = sec y; Can you sho me how to prove this?

Aug 05, 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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