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Trigonometry : Understanding Trigonometric Equations

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Example Questions

Example Question #6 : Factoring Trigonometric Equations

Factor the following expression

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

Possible Answers:

$\displaystyle (sin^nx+1)(sin^nx+2)$

$\displaystyle (sin^nx+1)(sin^nx-2)$

$\displaystyle (2sin^nx+1)(sin^nx-2)$

We cannot factor the above expression.

$\displaystyle (sin^nx-1)(sin^nx-2)$

Correct answer:

We cannot factor the above expression.

Explanation:

Letting $\displaystyle sin^n(x)=X$, we have the equivalent expression:

$\displaystyle X^2+2X+3$.

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

This shows that we cannot factor the above expression.

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Example Question #1 : Factoring Trigonometric Equations

Factor

$1-cos^{11}x$ $\displaystyle 1-cos^{11}x$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

We first note that we have:

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

Then taking $\displaystyle a=cos(x)$, we have the result.

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

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Example Question #4 : Factoring Trigonometric Equations

Find a simple expression for the following :

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

Possible Answers:

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

3sinx $\displaystyle 3sinx$

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

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

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

Correct answer:

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

Explanation:

First of all we know that :

$\displaystyle cos^2x+sin^2x=1$ and this gives:

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

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

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

we have then:

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

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Example Question #21 : Understanding Trigonometric Equations

What is a simple expression for the formula:

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

Possible Answers:

$\displaystyle (cosx-sinx)(2-cosx sinx)$

$\displaystyle (cosx+sinx)(1+cosx sinx)$

$\displaystyle (cosx-sinx)(1+3cosx sinx)$

$\displaystyle 2(cosx-sinx)(1+cosx sinx)$

$\displaystyle (cosx-sinx)(1+cosx sinx)$

Correct answer:

$\displaystyle (cosx-sinx)(1+cosx sinx)$

Explanation:

From the expression :

$\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$

we have:

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

Now since we know that :

$\displaystyle cos^2x+sin^2x=1$. This expression becomes:

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

This is what we need to show.

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Example Question #22 : Understanding Trigonometric Equations

Factor: $\displaystyle (sin^4x-cos^4x)$

Possible Answers:

$\displaystyle (sinx-cosx)(sinx+cosx)$

$\displaystyle (sin^2x-cos^2x)(sin^2x+cos^2x)$

$\displaystyle (sin^2x+cos^2x)(sinx+cosx)$

$\displaystyle -(sin^2x+cos^2x)(sinx-cosx)$

Correct answer:

$\displaystyle (sinx-cosx)(sinx+cosx)$

Explanation:

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

$\displaystyle x^4-y^4=(x^2+y^2)(x^2-y^2)=(x^2+y^2)(x-y)(x+y)$

Step 2: Factor the question:

$\displaystyle (sin^4x-cos^4x)=(sin^2x+cos^2x)(sin^2x-cos^2x)$

Factor more:

$\displaystyle (sin^2x+cos^2x)(sinx+cosx)(sinx-cosx)$

Step 3: Recall a trigonometric identity:

$\displaystyle sin^2x+cos^2x=1$.. Replace this

Final Answer: $\displaystyle (sinx+cosx)(sinx-cosx)$

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Trigonometry : Understanding Trigonometric Equations

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #6 : Factoring Trigonometric Equations

Example Question #1 : Factoring Trigonometric Equations

Example Question #4 : Factoring Trigonometric Equations

Example Question #21 : Understanding Trigonometric Equations

Example Question #22 : Understanding Trigonometric Equations

All Trigonometry Resources

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