Question about Super Tutor Trigonometry (ESDTRIG) for PC

Simple 0.5(cos15 + (squer of 3).sin 15) =0.5(cos15+(3x3).Sin15) =0.5(0.9659+9x0.2588) =0.5(0.9659+2.3292) =0.5x3.2951 =1.64755 Zulfikar Ali ali_zulfikar@yahoo.com 989980221

Posted on Mar 15, 2009

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Posted on Jan 02, 2017

It seems to me that you are trying to solve the quadratic equation

aX^2+bX+c=10 with a=-3, b=3, c=15 or**-3X^2+3X+15=0**.

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation**-3X^2+3X+15=0**..

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

**Solution1 =X_1=(-3-SQRT(189))/(-2*3)=(1+SQRT(21))/2**

**Solution2 =X_2=(-3+SQRT(189))/(-2*3)=(1-SQRT(21))/2** or about -1.791287847

Here SQRT stands for square root.

aX^2+bX+c=10 with a=-3, b=3, c=15 or

Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation

You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).

Discriminant is usually represented by the Greek letter DELTA (a triangle)

DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189

If the discriminant is positive (your case) the equation has two real solutions which are given by

Here SQRT stands for square root.

Aug 17, 2014 | Computers & Internet

tan(x/2)=sin(x)/(1+cos(x))

Setting x/2=45, means that x=90 (degrees)

But cos(90)=0 and sin(90)=1. Thus tan(45)=1/(1+0)=1.

Setting x/2=45, means that x=90 (degrees)

But cos(90)=0 and sin(90)=1. Thus tan(45)=1/(1+0)=1.

Mar 13, 2013 | SoftMath Algebrator - Algebra Homework...

Since X = pi, (3.141592...) I'm going to assume the problem is in radians not degrees.

Substitute pi in for X in the first equation and you get y = 4 sin(pi)

The sin of pi radians is zero.

Therefore, y = 4 * 0 = 0

I hope this helps you out.

Substitute pi in for X in the first equation and you get y = 4 sin(pi)

The sin of pi radians is zero.

Therefore, y = 4 * 0 = 0

I hope this helps you out.

Dec 15, 2010 | SoftMath Algebrator - Algebra Homework...

Sin 0 = 0

Sin 90 = 1

Sin 180 = 0

Sin 270 = -1

Sin 360 = Sin 0 = 0

Is there an angle greater than 360???

Is there any greater value of Sine than 1 ???

Sin 90 = 1

Sin 180 = 0

Sin 270 = -1

Sin 360 = Sin 0 = 0

Is there an angle greater than 360???

Is there any greater value of Sine than 1 ???

Aug 11, 2010 | Pyramid Calculus WIZ 2.0 (cp11395) for PC,...

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...

Hii..this can be done as follow

cos^2 x = 1 - sin^2 x--------(1)

2cos^2 x=1+ sin x

2cos^2 x - sin x -1=0

Substituting formula (1)

2(1 - sin^2 x) - sin x - 1 = 0

2sin^2 x + sin x - 1 = 0

Factor this

(2 sin x - 1)(sin x +1) = 0

2 sin x - 1 = 0 or sin x +1 = 0

sin x = 1, and sin x = -1

cos^2 x = 1 - sin^2 x--------(1)

2cos^2 x=1+ sin x

2cos^2 x - sin x -1=0

Substituting formula (1)

2(1 - sin^2 x) - sin x - 1 = 0

2sin^2 x + sin x - 1 = 0

Factor this

(2 sin x - 1)(sin x +1) = 0

2 sin x - 1 = 0 or sin x +1 = 0

sin x = 1, and sin x = -1

Apr 26, 2010 | SoftMath Algebrator - Algebra Homework...

we can represent 5 sin X --3 cos X as

5.83 sin ( X - 30.96)

if you want steps please leave a comment and don't forget to vote for me thank you

5.83 sin ( X - 30.96)

if you want steps please leave a comment and don't forget to vote for me thank you

Apr 14, 2010 | Super Tutor Trigonometry (ESDTRIG) for PC

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

Hi rowanwah

The sine of an angle is only applicable is a right triangle. If you just want a number, ie, the actual value of the sine 15 degrees you can look it up on Google. Do a search for "sine and cosine functions"

If you want the mathematical description of the sine of an angle it is described as follows

In a triangle ABC, there are 3 angles angle A, angle B and angle C. There are also 3 sides, Side AB, Side AC and side BC. The sine of angle A is equal to the side opposite Angle A divided by the Hypotenuse (the longest side opposite the right angle)

The Cosine of angle A is equal to the side adjacent to Angle A divided by the hypotenuse

Hope this helps Loringh PS Please leave a rating for me Thanks

The sine of an angle is only applicable is a right triangle. If you just want a number, ie, the actual value of the sine 15 degrees you can look it up on Google. Do a search for "sine and cosine functions"

If you want the mathematical description of the sine of an angle it is described as follows

In a triangle ABC, there are 3 angles angle A, angle B and angle C. There are also 3 sides, Side AB, Side AC and side BC. The sine of angle A is equal to the side opposite Angle A divided by the Hypotenuse (the longest side opposite the right angle)

The Cosine of angle A is equal to the side adjacent to Angle A divided by the hypotenuse

Hope this helps Loringh PS Please leave a rating for me Thanks

Nov 15, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

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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