Cos x + root 3 sin x =root 2

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

Posted on Jan 10, 2009

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Posted on Jan 02, 2017

Square each side

Sin ^2 (x) + cos ^2 (x) +2 sin (x)cos (x) = 49/25

1 + 2sin (x)cos (x) = 1.960

sin (2x) = 0.960

2x = 73.74 deg

x = 36.87 deg

Sin ^2 (x) + cos ^2 (x) +2 sin (x)cos (x) = 49/25

1 + 2sin (x)cos (x) = 1.960

sin (2x) = 0.960

2x = 73.74 deg

x = 36.87 deg

Sep 07, 2014 | Computers & Internet

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...

Press the relevant function key [SIN],[COS],[TAN] followed by the angle value.

For inverse trigonometric functions [SIN^-1], COS^-1], or [TAN^-1], Press [SHIFT][SIN], [SHIFT][COS], or [SHIFT][TAN].

For other functions

sec(x)=1/cos(x)

csc(x)=1/sin(x)

cot(x)=1/tan(x)

When calculating trigonometric functions one must make sure that the angle unit the calculator is using is the correct one.

For inverse trigonometric functions [SIN^-1], COS^-1], or [TAN^-1], Press [SHIFT][SIN], [SHIFT][COS], or [SHIFT][TAN].

For other functions

sec(x)=1/cos(x)

csc(x)=1/sin(x)

cot(x)=1/tan(x)

When calculating trigonometric functions one must make sure that the angle unit the calculator is using is the correct one.

Aug 26, 2010 | Casio FX-115ES Scientific Calculator

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...

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...

Hello,

Frankly my dear, who wants to compute the 99th derivative of sin(x)?

d/dx(sin(x))= cos(x)

d^2/dx^2 (sin(x))= d/dx( cos(x))=-sin(x)

d^3/dx^3 (sin(x))= -cos(x)

d^4/dx^4 (sin(x))=d/dx(-cos(x))=sinx.

In 100 successive differentiations you perform the foregoing sequence 100/4=25 times, and you end up with sin(x). Hence in the 99th step the derivative was -cos(x).

Hope it helps.

Frankly my dear, who wants to compute the 99th derivative of sin(x)?

d/dx(sin(x))= cos(x)

d^2/dx^2 (sin(x))= d/dx( cos(x))=-sin(x)

d^3/dx^3 (sin(x))= -cos(x)

d^4/dx^4 (sin(x))=d/dx(-cos(x))=sinx.

In 100 successive differentiations you perform the foregoing sequence 100/4=25 times, and you end up with sin(x). Hence in the 99th step the derivative was -cos(x).

Hope it helps.

Oct 20, 2009 | Casio FX-300MS 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

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

if you want to square sin(3), press:

(

sin

3

)

)

^

2

(

sin

3

)

)

^

2

Sep 24, 2007 | Texas Instruments TI-89 Calculator

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