Question about SoftMath Algebrator - Algebra Homework Solver (689076614429)

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).

Posted on Jan 22, 2012

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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

Press g [DEG] 5 5 SIN 4 5 COS * 1 * 1 * 1 *

I don't know why you need the three multiplies by 1. This could be done as 5 5 SIN 4 5 COS *

I don't know why you need the three multiplies by 1. This could be done as 5 5 SIN 4 5 COS *

Oct 30, 2012 | HP 15C Scientific Calculator

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 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...

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

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

You should buy a dog that takes you on walks.

Then you won't have the problem that you have.

Then you won't have the problem that you have.

Aug 23, 2009 | Microsoft Computers & Internet

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

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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