Question about SoftMath Algebrator - Algebra Homework Solver (689076614429)

# I need to rewrite Y=5(sqrt2)sin(x)-5(sqrt2)cos(x) as Y=Asin(Bx-c)

Posted by on

• Level 3:

An expert who has achieved level 3 by getting 1000 points

Superstar:

An expert that got 20 achievements.

All-Star:

An expert that got 10 achievements.

MVP:

An expert that got 5 achievements.

• SoftMath Master

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

Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).
goodluck!

Posted on Jan 02, 2017

×

my-video-file.mp4

×

## Related Questions:

### How can I solve Sinx+cosx=7/5?

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

Sep 07, 2014 | Computers & Internet

### Sinus calculation HP-15C

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 *

Oct 30, 2012 | HP 15C Scientific Calculator

### Find the trig function given its period cos 5 pi

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

Dec 12, 2011 | Super Tutor Trigonometry (ESDTRIG) for PC

### Differentiate each of the following w.r.t.x; 29.sin2xsinx

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
1. (29)'=0 derivative of a constant is zero
2. (sin(2X))'=cos(2X)*(2X)'=2*cos(2X)
3. (sin(X))'=cos(X)
Result is
(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...

### Please help me change: 5 sin X - 3 Cos X to C sin (X - @) @ = theta

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

5.83 sin ( X - 30.96)

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

### How do you use the sin-1

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

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

### Help

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

Feb 02, 2009 | Super Tutor Trigonometry (ESDTRIG) for PC

### Trig Identities

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.

May 22, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

### Cosh Squared or Cubed

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

(
sin
3
)
)
^
2

Sep 24, 2007 | Texas Instruments TI-89 Calculator

## Open Questions:

#### Related Topics:

141 people viewed this question

Level 3 Expert

Level 3 Expert