Question about Super Tutor Trigonometry (ESDTRIG) for PC

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Trig Identities cosecӨ-sinӨ=cos²Ө/sinӨ

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  • Anonymous May 19, 2009

    what are the primary operations used in trigonometry?was it like in geometry and advanced algebra?


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4 Answers

1/sin-sin
1-(sin*sin)/sin
(cos*cos)/sin

Posted on Sep 02, 2008

Cos^tan^=sin^

Posted on Aug 12, 2008

How to determine the area of a triagle abc, if a=110mm, b=80mm and c=120degrees

Posted on Jul 16, 2008

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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.

Posted on Jun 13, 2008

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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

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...

1 Answer

Sin x cos x = -1/2 solve for x


sin x cos x = -1/2
=> 2sinx cosx = -1
=> sin(2x) = -1
=> 2x = (3pi)/2 OR 2x = 270°
=> x = 3pi/4 OR x = 135°



Jan 04, 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)

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

(sin-cos)(sin+cos)


(sinX-cosX)(sinX+cosX) =(sin^2 - Cos^2X)
=Sin^2X - (1-Sin^2X)
=Sin^2X -1 + Sin^2X
=2Sin^2X -1

Zulfikar Ali
ali_zulfikar@yahoo.com
9899780221

Jan 26, 2009 | Super Tutor Trigonometry (ESDTRIG) for PC

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

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