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Featured 6 months ago

Strategy: Rewrite this equation as a quadratic equation using

Step 1. Rewrite this equation as a quadratic using

You are given

Replace

Step 2. Factor the quadratic equation.

Solving gives us

Step 3. Replace

These answer work because we were asked to find the solutions in

However, the graph of

Which shows **four** solutions in the interval

We must find the other solutions by recognizing that sine functions are periodic with respect to

So,

The other two solutions are

Featured 4 months ago

Let

- now we've got a quadratic equation that can be solved by using the quadratic formula
#u=(-b+-sqrt(b^2-4ac))/(2a)# .

we can get the values the values of#a, b, and c# by comparing our equation to the standard form#au^2+bu+c# , and the values are#a=1, b=-1, c=-6#

#u=(-b+-sqrt(b^2-4ac))/(2a)=>tantheta=(-b+-sqrt(b^2-4ac))/(2a)#

#=(-(-1)+-sqrt((-1)^2-4(1)(-6)))/(2(1))#

#tantheta=(1+5)/2" or "tantheta=(1-5)/2#

#tantheta=3" or "tantheta=-2#

- a period of the function
#tantheta# is#pi# therefore#tan(theta+pi)=tantheta#

Let

#tantheta_1=3" and " tantheta_2=-2#

#theta_1=tan^-1(3)+npi; "where "n inZZ#

#theta_2=tan^-1(-2)+npi" where "n in ZZ#

Featured 4 months ago

One does not use integration or rationalizing to prove this. One proves this by changing only 1 side of the equation, until it is identical to the other side.

Prove:

The identity for

Use the fact that

Write

Multiply the fraction inside the square by 1 in the form of

Perform the multiplication:

Please observe the cancellation of the embedded denominators:

Write the equation without the cancelled factors:

Perform the multiplication implied by the square:

Use the identity

Use the identity

Featured 3 months ago

In order to answer this I have assumed a vertical shift of

The standard cos function

If we want a period of

That is

To get an amplitude of

There is to be no horizontal shift, so the argument for

In order to achieve the vertical shift (that I assumed would be

Featured 2 months ago

See below

There are several definitions of the function

(i) Probably the simplest is in relation to any right

So, in the case above

(ii) Consider the unit circle centred at the origin

NB: This definition is equivalent to (i) above for

(iii)

Hence, it can be defined by a Taylor series at

Finally, to "represent"

graph{sinx [-6.244, 6.244, -3.12, 3.124]}

Featured 2 weeks ago

The way that one proves an identity is to make substitutions to only one side until it is identical to the other side:

Given:

Use the property

Digress and prove that

An alternate form for

Substitute

An alternate form for

Distribute through

Distribute through

use the property

Write the

Factor out

Use the property of logarithms

Use the property

When we make a common denominator we obtain:

Combine over the common denominator:

The leading coefficient multiplied into the numerator becomes 1:

Use the identity

Substitute this into equation [1]:

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