Factoring Trigonometric Equations

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Trigonometry › Factoring Trigonometric Equations

Questions 1 - 10

1

Factor $1-2\cos x+\cos^2 x$ .

$(1-\cos x)^2$

$\sin^2x$

$\sin^4 x$

$\cos^4 x$

$\sin x \tan x$

Explanation

Don't get scared off by the fact we're doing trig functions! Factor as you normally would. Because our middle term is negative ( $-2\cos x$ ), we know that the signs inside of our parentheses will be negative.

This means that $1-2\cos x+\cos^2 x$ can be factored to $(1-\cos x)(1-\cos x)$ or $(1-\cos x)^2$ .

2

Which of the following values of in radians satisfy the equation

$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\frac{3\pi}{4}$

1 only

2 only

3 only

1 and 2

1, 2, and 3

Explanation

The fastest way to solve this equation is to simply try the three answers. Plugging in $\frac{\pi}{4}$ gives

$tan^2(\frac{\pi}{4})-tan(\frac{\pi}{4})=(1)^2-1=1-1=0$

Our first choice is valid.

Plugging in $\frac{\pi}{2}$ gives

$tan^2(\frac{\pi}{2})-tan(\frac{\pi}{2})$

However, since $tan(\frac{\pi}{}2)$ is undefined, this cannot be a valid answer.

Finally, plugging in $\frac{3\pi}{4}$ gives

$tan^2(\frac{3\pi}{4})-tan(\frac{3\pi}{4})=(-1)^2-(-1)=1+1=2$

Therefore, our third answer choice is not correct, meaning only 1 is correct.

3

$f(x)=\sin^2(x)+\cos^2(x)-\sin(2x)$

Find the zeros of the above equation in the interval

$[0,\pi]$ .

$\frac{\pi}{4}$

$\pi$

$\frac{\pi}{2}$

$2\pi$

Explanation

$f(x)=\sin^2(x)-\sin(2x)+\cos^2(x)=1-2\sin(x)cos(x)$

$=1-2\sin(x)cos(x)=0$

Therefore,

$\frac{1}{2}=sin(x)cos(x)$

and that only happens once in the given interval, at $\frac{\pi}{4}$ , or 45 degrees.

$\sin\bigg(\frac{\pi}{4}\bigg)\cos\bigg(\frac{\pi}{4}\bigg)=\frac{1}{\sqrt2}\cdot \frac{1}{\sqrt2}=\frac{1}{2}$

4

Factor the expression

$(cos(x)+sin(x))^{2}$

$2(cos(x)+sin(x))^{2}$

$(cos(x)-sin(x))^{2}$

$(cos(x)+2sin(x))^{2}$

$(cos(x)+sin(x))^{3}$

Explanation

We have .

Now since

This last expression can be written as :

.

This shows the required result.

5

Factor the following expression:

$sin^{3}x-1$

Explanation

We know that we can write

in the following form

.

Now taking ,

we have:

.

This is the result that we need.

6

We accept that :

What is a simple expression of

$cos^{4}x-sin^{4}x$

Explanation

First we see that :

.

Now letting

we have

We know that :

and we are given that

, this gives

7

Factor the following expression:

We can't factor this expression.

Explanation

Note first that:

and :

.

Now taking $a=cosx \quad b=sinx$ . We have

.

Since and .

We therefore have :

8

Factor the following expression

$sin^{2n}(x)+2sin^n(x)+3$ where is assumed to be a positive integer.

We cannot factor the above expression.

Explanation

Letting , we have the equivalent expression:

.

We cant factor since $X^2+2X+3=(x+1)^2+2 >0 \forall \quadx\in\mathbb{R}$ .

This shows that we cannot factor the above expression.

9

Factor

$1-cos^{11}x$

$(1-cosx)(cos^{10}x+cos^{9}x+...cosx+1)$

$(1-cosx)(cos^{10}x+cos^{9}x+...cosx-1)$

$(-1-cosx)(cos^{10}x+cos^{9}x+...cosx+1)$

$(1-cosx)(cos^{11}x+cos^{9}x+...cosx+1)$

$(2-cosx)(cos^{10}x+cos^{9}x+...cosx+1)$

Explanation

We first note that we have:

$1-a^{11}=(1-a)(a^{10}+a^9+...a+1)$

Then taking , we have the result.

$(1-cosx)(cos^{10}x+cos^{9}x+...cosx+1)$

10

Find a simple expression for the following :

$cos^{2n}x-cos^{2n+2} x$

$cos^{2n}xsin^{2}x$

$cos^{2n}x (1-sin^{2}x)$

$-cos^{2n}xsin^{2}x$

$2cos^{2n}xsin^{2}x$

Explanation

First of all we know that :

and this gives:

$1-cos^2{x}=sin^{2}x$ .

Now we need to see that: $cos^{2n}x-cos^{2n+2} x$ can be written as

$cos^{2n}x(1-cos^2x)$ and since $1-cos^2{x}=sin^{2}x$

we have then:

$cos^{2n}x(1-cos^2x)=cos^{2n}xsin^2x$ .

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