Understanding Trigonometric Equations - Trigonometry

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Question

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

Answer

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 $(1-\cos x)^2$ .

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Question

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

Find the zeros of the above equation in the interval

$[0,\pi]$ .

Answer

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

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

Therefore,

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

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

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

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Question

Which of the following values of in radians satisfy the equation

$\frac{\pi}{4}$

$\frac{\pi}{2}$

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

Answer

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.

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Question

Factor the following expression:

Answer

Note first that:

and :

.

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

.

Since and .

We therefore have :

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Question

Factor the expression

Answer

We have .

Now since

This last expression can be written as :

.

This shows the required result.

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Question

Factor the following expression

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

Answer

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.

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Question

Factor

$1-cos^{11}x$

Answer

We first note that we have:

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

Then taking , we have the result.

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

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Question

Find a simple expression for the following :

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

Answer

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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Question

Factor the following expression:

$sin^{3}x-1$

Answer

We know that we can write

in the following form

.

Now taking ,

we have:

.

This is the result that we need.

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Question

What is a simple expression for the formula:

$cos^{3}x-sin^{3}x$

Answer

From the expression :

we have:

$cos^{3}x-sin^{3}x=(cosx-sinx)(cos^2x+cosxsinx+sin^2x)$

Now since we know that :

. This expression becomes:

$cos^{3}x-sin^{3}x=(cosx-sinx)(1+cosxsinx)$ .

This is what we need to show.

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Question

We accept that :

What is a simple expression of

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

Answer

First we see that :

.

Now letting

we have

We know that :

and we are given that

, this gives

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Question

Factor:

Answer

Step 1: Recall the difference of squares (or powers of four) formula:

Step 2: Factor the question:

Factor more:

Step 3: Recall a trigonometric identity:

.. Replace this

Final Answer:

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Question

A triangle has sides , , of lengths , , respectively. The angle opposite each side is called , , , respectively. The sine of which angle and the cosine and which different angle will both yield $\frac{8}{17}$

In the answer, list the sine first and the cosine second.

Answer

This is the figure being described in the problem, and as sine is opposite over hypotenuse and cosine adjacent over hypotenuse, the sine of and the cosine of will yield the correct answer.

$\sin(a)=\frac{opp}{hyp}=\frac{8}{17}$

$\cos(b)=\frac{adj}{hyp}=\frac{8}{17}$

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Question

Solve the equation over the domain (answer in degrees).

Answer

Rearrange algebraically so that,

.

Over the interval 0 to 360 degrees, sinx = 1 when x equals 90 degrees.

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Question

Solve each equation over the domain (answer in degrees).

Answer

First, think of the angle values for which . (This is the equivalent of taking the arctan.)

$x-15=tan^{-1}(1)$

$x=tan^{-1}+15$ .

The angles for which this is true are 45 degrees and 225 degrees.

We set x-15 equal to those two angles and solve for x, giving us 60 and 240.

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Question

Solve each equation over the domain (answer in degrees).

Answer

Rearrange the equation so that,

.

Recall the angles over the interval 0 to 360 degrees for which sec is equal to 2.

These are 60 and 300 degrees.

Set x+7 equal to these angle measures and then find that x equals 53 and 293.

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Question

Solve each equation over the interval $2\pi$

Answer

Rearrange the equation so that,

$cos^2x = \frac{1}{2}$ .

Take the square of both sides and then recall the angle measures for which,

$cos(x)={\frac{1}{\sqrt2}}$ .

These measures over the interval $0<x< 2 \pi$

$x=\frac{\pi}{4}; \frac{7\pi}{4}$ .

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Question

Solve each quation over the interval $0 < x < 2\pi$

Answer

Rearrange the equation so that,

.

Take the square of both sides and find the angles for which

$sin(x)=\pm1$ .

These two angles are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ .

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Question

Solve for using trigonometric ratios.

Answer

To solve for x, first set up a trigonometric equation using the information provided in the diagram. The two side lengths given are the hypotenuse, x, and the side opposite the given angle, 6. We can set up our equation like this:

$sin\left(\frac{\pi}{4}\right)= \frac{6}{x}$

The sine of $\frac{\pi}{4}$ is $\frac{1}{\sqrt2}$ , so we can substitue that in:

$\frac{1}{\sqrt2}=\frac{6}{x}$

cross multiplying gives us $x=6\sqrt2$ .

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Question

Solve for using trigonometric ratios.

Answer

To solve for x, first set up a trigonometric equation using the information provided in the diagram. The two side lengths given are x, the side opposite the angle, and 3, the side adjacent to the angle. This means we'll be using tangent. Set up the equation like this:

$tan\left(\frac{\pi}{3}\right)=\frac{x}{3}$

We can't just know the tangent by using the unit circle, but we can easily figure it out using sine and cosine. Tangent can be evaluated as sine over cosine.

The sine of $\frac{\pi}{3}$ is $\frac{\sqrt3}{2}$ , and the cosine is $\frac{1}{2}$ . Find the tangent by dividing:

$tan\left(\frac{\pi}{3}\right)=\frac{\sqrt3}{2}\div \frac{1}{2}$

$tan\left(\frac{\pi}{3}\right)=\frac{\sqrt3}{2}\cdot \frac{2}{1}$

$tan\left(\frac{\pi}{3}\right)=\sqrt3$

Now we can substitute that value into the original equation we set up:

$\sqrt3 = \frac{x}{3}$ multiply both sides by 3

$3\sqrt3=x$

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