Question about Super Tutor Trigonometry (ESDTRIG) for PC

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

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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Posted on Jan 02, 2017

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...

you can follow this link and can get the solution http://mathforum.org/dr.math/faq/formulas/faq.trig.html

May 31, 2010 | Super Tutor Trigonometry (ESDTRIG) for PC

sin x cos x = -1/2

=> 2sinx cosx = -1

=> sin(2x) = -1

=> 2x = (3pi)/2 OR 2x = 270°

=> x = 3pi/4 OR x = 135°

=> 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...

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

(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

=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

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

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

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