Featured 1 month ago

Note that

In this case

"Which angle has a

The only way to determine this is with a calculator or tables.

Using a graph is possible, but not accurate enough.

Depending on which calculator you have, the following are possible key presses:

D.A.L. calculators:

shift

The answer will be

Or

The answer will be

Note that this is only the acute angle, the 1st quadrant angle.

In the 4th quadrant,

Featured 1 month ago

If

then

and

If

then

and

Plugging in the values for

If we call this value

then

and (since

Featured 4 days ago

Use the cosine angle addition formula:

#cos(x+y)=cosxcosy-sinxsiny#

We need to determine

**Determining**

#tan^2x+1=sec^2x" "" "" "# use#tanx=5/3#

#25/9+1=sec^2x#

#secx=sqrt(34/9)=sqrt34/3#

Then:

#color(blue)(cosx=1/secx=3/sqrt34#

Now using:

#sin^2x+cos^2x=1" "" "" "# where#cosx=3/sqrt34#

#sin^2x+9/34=1#

#color(blue)(sinx=sqrt(25/34)=5/sqrt34#

**Determining**

#sin^2y+cos^2y=1" "" "" "# use#siny=1/3#

#1/9+cos^2y=1#

#color(blue)(cosy=sqrt(8/9)=(2sqrt2)/3#

**Returning to**

#cos(x+y)=cosxcosy-sinxsiny#

#color(white)(cos(x+y))=3/sqrt34((2sqrt2)/3)-5/sqrt34(1/3)#

#color(white)(cos(x+y))=(6sqrt2-5)/(3sqrt34)#

Note these are all assuming that

Featured 4 days ago

The explanation is written with the assumption that you meant

Use the distributive property :

Substitute x for

Square both sides:

Add

Remove a factor of the 3 from the first 3 terms:

Use the middle in the right side of the pattern

Substitute the left side of the pattern into the equation:

Substitute

Divide both sides by

Write the denominators as squares:

The is standard Cartesian form of the equation of an ellipse with a center at

Featured 1 week ago

First use the cosine angle-addition formula:

#cos(a+b)=cos(a)cos(b)-sin(a)sin(b)#

Then the original expression equals:

#=cos(tan^-1(2))cos(tan^-1(3))-sin(tan^-1(2))sin(tan^-1(3))#

Note that if

So, when

#cos(tan^-1(2))=cos(theta)="adjacent"/"hypotenuse"=1/sqrt5#

#sin(tan^-1(2))=sin(theta)="opposite"/"hypotenuse"=2/sqrt5#

Now let angle

#{("opposite"=3),("adjacent"=1),("hypotenuse"=sqrt10):}#

Then:

#cos(tan^-1(3))=cos(phi)="adjacent"/"hypotenuse"=1/sqrt10#

#sin(tan^-1(3))=sin(phi)="opposite"/"hypotenuse"=3/sqrt10#

Plugging these into the expression from earlier we get:

#=1/sqrt5(1/sqrt10)-2/sqrt5(3/sqrt10)#

Note that

#=(1-6)/(5sqrt2)=-1/sqrt2#

Featured 4 days ago

Given

To get x-intercepts we can put y=0 and solve for

So

Putting

Putting

Putting

Putting

(a) So points of x-intercepts over

b) For maximum value of y the value of

So

For

The point(in (0,2pi) where the graph of this function reaches

maximum is

Featured 4 days ago

Start from a "basic cycle" for the

Starting with the "basic cycle" for

Then consider what values of

(Yes; I know: because

Giving us the "basic cycle" for

Adding

In the image below, I have added the "non-basic cycles" as well as the "basic cycle" used for analysis:

Featured 4 days ago

See explanation...

Consider a right angled triangle with an internal angle

Then:

#sin theta = a/c#

#cos theta = b/c#

So:

#sin^2 theta + cos^ theta = a^2/c^2+b^2/c^2 = (a^2+b^2)/c^2#

By Pythagoras

So given Pythagoras, that proves the identity for

For angles outside that range we can use:

#sin (theta + pi) = -sin (theta)#

#cos (theta + pi) = -cos (theta)#

#sin (- theta) = - sin(theta)#

#cos (- theta) = cos(theta)#

So for example:

#sin^2 (theta + pi) + cos^2 (theta + pi) = (-sin theta)^2 + (-cos theta)^2 = sin^2 theta + cos^2 theta = 1#

**Pythagoras theorem**

Given a right angled triangle with sides

The area of the large square is

The area of the small, tilted square is

The area of each triangle is

So we have:

#(a+b)^2 = c^2 + 4 * 1/2ab#

That is:

#a^2+2ab+b^2 = c^2+2ab#

Subtract

#a^2+b^2 = c^2#

Featured 4 days ago

Use the formula for a circle

The formula for a circle centred at the origin is

#x^2+y^2=r^2#

That is, the distance from the origin to any point

Picture a circle of radius

graph{(x^2+y^2-1)((x-sqrt(3)/2)^2+(y-0.5)^2-0.003)=0 [-2.5, 2.5, -1.25, 1.25]}

If we draw a line from that point to the origin, its length is

graph{(x^2+y^2-1)(y-sqrt(3)x/3)((y-0.25)^4/0.18+(x-sqrt(3)/2)^4/0.000001-0.02)(y^4/0.00001+(x-sqrt(3)/4)^4/2.7-0.01)=0 [-2.5, 2.5, -1.25, 1.25]}

Let the angle at the origin be theta (

Now for the trigonometry.

For an angle

#sin theta = "opp"/"hyp" = y/r" "<=>" "y=rsintheta#

Similarly,

#cos theta = "adj"/"hyp"=x/r" "<=>" "x = rcostheta#

So we have

#" "x^2" "+" "y^2" "=r^2#

#(rcostheta)^2+(rsintheta) ^2 = r^2#

#r^2cos^2theta + r^2 sin^2 theta = r^2#

The

#cos^2 theta + sin^2 theta = 1#

This is often rewritten with the

#sin^2 theta + cos^2 theta = 1#

And that's it. That's really all there is to it. Just as the distance between the origin and any point