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Precalculus : Pre-Calculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #145 : Trigonometric Functions

Compute cos(4x) in function of sin(x) .

Possible Answers:

1-8sin^2(x)-8sin^4(x)

1+8sin^2(x)+8sin^4(x)

1-8sin^2(x)+8sin^4(x)

1-6sin^2(x)+8sin^4(x)

1-8sin^2(x)+6sin^4(x)

Correct answer:

1-8sin^2(x)+8sin^4(x)

Explanation:

Using trigonometric identities we have :

cos(4x)=1-2sin^2(2x) and we know that:

sin(2x)=2sin(x)cox(x)

This gives us :

cos(4x)=1-2(2sin(x)cox(x))^2=1-8sin^2(x)cos^2(x)

= 1-8sin^2(x)(1-sin^2(x)= 1-8sin^2(x)+8sin^4(x)

Hence:

cos(4x)=1-8sin^2(x)+8sin^4(x)

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Example Question #31 : Fundamental Trigonometric Identities

Given that :

$tan(2x)=\frac{2tan(x)}{1-tan^2(x)}$

Let,

$tan\left(\frac{x}{8}\right)=s$

What is $tan\left(\frac{x}{4}\right)$ in function of ?

Possible Answers:

$\frac{4s}{1-4s^2}$

$\frac{s}{1-4s^2}$

$\frac{2s}{1-s^2}$

$\frac{4s}{1-s^2}$

$\frac{2s}{1-2s^2}$

Correct answer:

$\frac{2s}{1-s^2}$

Explanation:

We will use the given formula :

We have in this case:

$tan\left(\frac{x}{4}\right)=\frac{2tan(\frac{x}{8})}{1-tan^2(\frac{x}{8})}$

Since we know that :

$tan\left(\frac{x}{8}\right)=s$

This gives :

$\frac{2tan(\frac{x}{8})}{1-tan^2(\frac{x}{8})}=\frac{2s}{1-s^2}$

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Example Question #1511 : Pre Calculus

Using the fact that cos(x+180)=-cos(x) , what is the result of the following sum:

$cos(1)+cos(2)+cos(3)\cdots +cos(359)$

Possible Answers:

Correct answer:

Explanation:

We can write the above sum as :

$cos(1)+cos(181)+cos(2)+cos(182)\cdots + cos(179)+cos(379) + cos(180)$

From the given fact, we have :

cos(1)+cos(181)=cos(1)+cos(1+180)=cos(1)-cos(1)=0

cos(2)+cos(182)=cos(2)+cos(2+180)=cos(2)-cos(2)=0

$\vdots$

cos(179)+cos(379)=cos(179)+cos(179+180)=cos(179)-cos(179)=0

This gives us :

$cos(1)+cos(181)+cos(2)+cos(182)\cdots + cos(179)+cos(379) + cos(180)$

=0+cos(180)=-1

Therefore we have:

$cos(1)+cos(2)+cos(3)\cdots +cos(359)=-1$

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Example Question #147 : Trigonometric Functions

Let $\alpha, \beta$ be real numbers. If $tan(2x)=\alpha$ and $tan(x)=\beta$

What is the value of $\alpha$ in function of $\beta$ ?

Possible Answers:

$\alpha=\frac{2\beta}{1-\beta^2}$

$\alpha=\frac{2\beta^2}{1-\beta^2}$

$\alpha=\frac{2\beta}{1-2\beta^2}$

$\alpha=\frac{2\beta}{2-\beta^2}$

$\alpha=\frac{2\beta}{3-2\beta^2}$

Correct answer:

$\alpha=\frac{2\beta}{1-\beta^2}$

Explanation:

Using trigonometric identities we know that :

$tan(2x)=\frac{2tan(x)}{1-tan^2(x)}$

This gives :

$tan(2x)=\frac{2\beta}{1-\beta^2}$

We also know that $\alpha =tan(2x)$

This gives :

$\alpha=\frac{2\beta}{1-\beta^2}$

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Example Question #31 : Fundamental Trigonometric Identities

Given that :

cos(a+b)=cos(a)cos(b)-sin(a)sin(b) and,

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

Compute :

cos(3a) in function of cos(a) .

Possible Answers:

3cos^3(a)-4cos(a)

-4cos^3(a)-3cos(a)

4cos^3(a)-3cos(a)

3cos^2(a)-4cos(a)

4cos^3(a)+3cos(a)

Correct answer:

4cos^3(a)-3cos(a)

Explanation:

We have using the given result:

cos(3a)=cos(2a)cos(a)-sin(2a)sin(a)

cos(2a)=cos^2(a)-sin^2(a)=2cos^2(a)-1

sin(2a)=2sin(a)cos(a)

This gives us:

cos(3a)=(2cos^2(a)-1)cos(a)-2cos(a)sin^2(a)

=2cos^3(a)-cos(a)-2cos(a)+2cos^3(a)=4cos^3(a)-3cos(a)

Hence :

cos(3a)=4cos^3(a)-3cos(a)

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Example Question #32 : Fundamental Trigonometric Identities

Let be an integer and a real number. Compute $cos(x+n\pi)$ as a function of cos(x) .

Possible Answers:

-(-1)^n cos(x)

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

- cos(x)

(-1)^n cos(x)

$(-1)^{n+1} cos(x)$

Correct answer:

(-1)^n cos(x)

Explanation:

Using trigonometric identities we have :

$cos(x+n\pi)=cos(x)cos(n\pi)-sin(x)sin(n\pi)$

We know that :

$sin(n\pi)=0$ and $cos(n\pi)=(-1)^n$

This gives :

$cos(x+n\pi)=(-1)^ncos(x)$

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Example Question #1511 : Pre Calculus

Compute $sin\left(\frac{\pi}{2}+n\pi\right)$ .

Possible Answers:

sin(1)

(-1)^n

Correct answer:

(-1)^n

Explanation:

Using trigonometric identities we know that:

sin(x+y)=sin(x)cos(y)+cos(x)sin(y)

Letting $x=\frac{\pi}{2}$ and $y=n\pi$ in the above expression we have:

$sin\left(\frac{\pi}{2}+n\pi\right)=sin\left(\frac{\pi}{2}\right)cos(n\pi)+sin(n\pi)cos\left(\frac{\pi}{2}\right)$

We also know that:

$cos(n\pi)=(-1)^n$ and $sin(n\pi)=0$ .

This gives:

$sin\left(\frac{\pi}{2}+n\pi\right)=(-1)^n$

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Example Question #34 : Fundamental Trigonometric Identities

Given that:

tan(x)=s , what is the value of $tan\left(x+\frac{\pi}{4}\right)$ in function of ?

Possible Answers:

$\frac{2+s}{2-s}$

$\frac{1+s}{1-s}$

$\frac{1+s}{1-s^2}$

$\frac{1+s^2}{1-s}$

$\frac{1+2s}{1-2s}$

Correct answer:

$\frac{1+s}{1-s}$

Explanation:

We know by definition that:

$tan\left(x+\frac{\pi}{4}\right)=\frac{sin(x+\frac{\pi}{4})}{cos(x+\frac{\pi}{4})}$

We also have by trigonometric identities:

$sin\left(x+\frac{\pi}{4}\right)=sin(x)cos\left(\frac{\pi}{4}\right)+sin\left(\frac{\pi}{4}\right)cos(x)=\frac{\sqrt{2}}{2}(sin(x)+cos(x))$

$cos\left(x+\frac{\pi}{4}\right)=cos(x)cos\left(\frac{\pi}{4}\right)-sin\left(\frac{\pi}{4}\right)sin(x)=\frac{\sqrt{2}}{2}(cos(x)-sin(x)) $$$

Thus :

$tan\left(x+\frac{\pi}{4}\right)=\frac{cos(x)+sin(x)}{cos(x)-sin(x)}$

Now we have:

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

This gives us:

$tan\left(x+\frac{\pi}{4}\right)=\frac{(1+\frac{sin(x)}{cos(x)})}{{(1-\frac{sin(x)}{cos(x)}})}=\frac{1+s}{1-s}$

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Example Question #41 : Fundamental Trigonometric Identities

Given that:

tan(x)=s

Compute

tan(2x) in function of .

Possible Answers:

$\frac{s(1-s^2)}{(1-s^2-2s)(1-s^2+2s)}$

$\frac{4s^2(1-s^2)}{(1-s^2-2s)(1-s^2+2s)}$

$\frac{s(1-s^2)}{(1-3s^2)(1+s^2)}$

$\frac{4s(1-s^2)}{(1+s^2-2s)(1-s^2+2s)}$

$\frac{4s-4s^3}{1-6s^2+s^4}$

Correct answer:

$\frac{4s-4s^3}{1-6s^2+s^4}$

Explanation:

We will use the following formulas for calculating the tangent:

$tan(2x)=\frac{2tan(x)}{1-tan^2(x)}$

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

We have then:

$tan(4x)=\frac{2tan(2x)}{1-tan^2(2x)}= 2\frac{\frac{2tan(x)}{1-tan^2(x)}}{1-\left(\left(\frac{2tan(x)}{1-tan^2(x)}\right)^2\right)}$

This gives us :

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

Simplifying a bit we get :

$\frac{4tan(x)}{(1-tan^2(x))} \cdot \frac{(1-tan^2(x))^2}{((1-tan^2(x))^2-4tan^2(x))}$

Writing

((1-tan^2(x))^2-4tan^2(x))=1-6tan^2x+tan^4x

We now have :

$tan(4x)=\frac{4tan(x)(1-tan^2x)}{1-6tan^2x+tan^4x}\rightarrow \frac{4tan(x)-4tan^3(x)}{1-6tan^2x+tan^4x}$

Using the fact that tan(x)=s , we have finally:

$tan(4x)= \frac{4s-4s^3}{1-6s^2+s^4}$

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Example Question #1512 : Pre Calculus

Which of the following expressions best represent cos(x)sin(x) ?

Possible Answers:

$\frac{cos(2x)}{2}$

$\frac{sin(x)+1}{2}$

$\frac{cos(2x)+1}{2}$

$\frac{sin(2x)}{2}$

$\frac{sin(2x)+1}{2}$

Correct answer:

$\frac{sin(2x)}{2}$

Explanation:

Write the trigonometric product and sum identity for cos(A)sin(B) .

$cos(A)sin(B)=\frac{1}{2}[sin(A+B)-sin(A-B)]$

For cos(x)sin(x) , replace A,B with and simplify the expression.

$\frac{1}{2}[sin(x+x)-sin(x-x)]=\frac{1}{2}[sin(2x)-sin(0)]$

$\frac{sin(2x)}{2}$

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #145 : Trigonometric Functions

Example Question #31 : Fundamental Trigonometric Identities

Example Question #1511 : Pre Calculus

Example Question #147 : Trigonometric Functions

Example Question #31 : Fundamental Trigonometric Identities

Example Question #32 : Fundamental Trigonometric Identities

Example Question #1511 : Pre Calculus

Example Question #34 : Fundamental Trigonometric Identities

Example Question #41 : Fundamental Trigonometric Identities

Example Question #1512 : Pre Calculus

All Precalculus Resources

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