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ACT Math : How to find negative sine

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Example Questions

Example Question #1 : Trigonometry

If sin(x)=0.5 , what is sin(8x) if is between and $2\pi$ ?

Possible Answers:

-0.5

-1.5

$-\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2}$

0.5

Correct answer:

$-\frac{\sqrt{3}}{2}$

Explanation:

Recall that $sin(\frac{\pi}{6}) =0.5$ .

Therefore, we are looking for $sin(8*\frac{\pi}{6})$ or $sin(\frac{4\pi}{3})$ .

Now, this has a reference angle of $\frac{\pi}{3}$ , but it is in the third quadrant. This means that the value will be negative. The value of $sin(\frac{\pi}{3})$ is $\frac{\sqrt{3}}{2}$ . However, given the quadrant of our angle, it will be $-\frac{\sqrt{3}}{2}$ .

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Example Question #61 : Trigonometry

What is the sine of the angle formed between the origin and the point (4,-10) if that angle is formed with one side of the angle beginning on the -axis and then rotating counter-clockwise to (4,-10) ?

Possible Answers:

$\frac{5}{\sqrt{29}}$

$-\frac{5}{\sqrt{29}}$

$4\sqrt{10}$

-2.5

2.5

Correct answer:

$-\frac{5}{\sqrt{29}}$

Explanation:

You can begin by imagining a little triangle in the fourth quadrant to find your reference angle. It would look like this:

Sin410

Now, to find the sine of that angle, you will need to find the hypotenuse of the triangle. Using the Pythagorean Theorem, a^2+b^2=c^2 , where and are leg lengths and is the length of the hypotenuse, the hypotenuse is $\sqrt{a^2+b^2}$ , or, for our data:

$\sqrt{4^2+10^2}=\sqrt{16+100}=\sqrt{116}=2\sqrt{29}$

The sine of an angle is:

$\frac{opposite}{hypotenuse}$

For our data, this is:

$\frac{10}{2\sqrt{29}}=\frac{5}{\sqrt{29}}$

Since this is in the fourth quadrant, it is negative, because sine is negative in that quadrant.

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Example Question #62 : Trigonometry

What is the sine of the angle formed between the origin and the point (-3,-8) if that angle is formed with one side of the angle beginning on the -axis and then rotating counter-clockwise to (-3,-8) ?

Possible Answers:

$\frac{8}{3}$

$-4\sqrt{55}$

$-\frac{3}{8}$

$\frac{8}{\sqrt{73}}$

$-\frac{8}{\sqrt{73}}$

Correct answer:

$\frac{8}{\sqrt{73}}$

Explanation:

You can begin by imagining a little triangle in the third quadrant to find your reference angle. It would look like this:

Sin38

$\sqrt{3^2+8^2}=\sqrt{9+64}=\sqrt{73}$

The sine of an angle is:

$\frac{opposite}{hypotenuse}$

For our data, this is:

$\frac{8}{\sqrt{73}}$

Since this is in the third quadrant, it is negative, because sine is negative in that quadrant.

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Example Question #63 : Trigonometry

If $sin(2x)=\frac{-\sqrt{2}}{2}$ , what is the value of if $\frac{\pi}{2} \le 2x \le \frac{3\pi}{2}$ ?

Possible Answers:

$\frac{\pi}{4}$

$\frac{5\pi}{4}$

$\pi$

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

$\frac{5\pi}{8}$

Correct answer:

$\frac{5\pi}{8}$

Explanation:

Recall that the 45-45-90 triangle appears as follows in radians:

454590rad

Now, the sine of $\frac{\pi}{4}$ is $\frac{1}{\sqrt{2}}$ . However, if you rationalize the denominator, you get:

$\frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

Since $sin(2x)=\frac{-\sqrt{2}}{2}$ , we know that must be represent an angle in the third quadrant (where the sine function is negative). Adding its reference angle to $\pi$ , we get:

$\pi + \frac{\pi}{4}=\frac{5\pi}{4}$

Therefore, we know that:

$2x = \frac{5\pi}{4}$ , meaning that $x=\frac{5\pi}{8}$

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ACT Math : How to find negative sine

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : Trigonometry

Example Question #61 : Trigonometry

Example Question #62 : Trigonometry

Example Question #63 : Trigonometry

All ACT Math Resources

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