Question about Casio FX-115ES Scientific Calculator

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Posted on Jan 02, 2017

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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...

From the given data we can get angle A=30 degrees and B = 30 degrees.

There fore

sin ( A+B) = sin ( 30+30) =sin 60 = root ( 3 )/ 2

There fore

sin ( A+B) = sin ( 30+30) =sin 60 = root ( 3 )/ 2

May 19, 2010 | Mathsoft Solving and Optimization...

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...

You would press on the SIN key, "sin(" should come on the display, then enter the value, for example 3.5 or -1 so that the display shows

sin(3,5) or sin(-1) for example.

remember to close the sin with the ) on the end.

**sin(-1) = -0.841470985**

sin(3,5) or sin(-1) for example.

remember to close the sin with the ) on the end.

Feb 16, 2010 | Texas Instruments TI-30XA Calculator

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

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

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