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Trigonometry : Trigonometric Identities

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Example Question #51 : Trigonometric Identities

Find $\sin 2\theta$ if $\cos \theta = \frac{4}{5}$ and $\frac{3\pi}{2}<\theta<2\pi$ .

Possible Answers:

$-\frac{3}{5}$

$-\frac{24}{25}$

$-\frac{12}{25}$

$\frac{24}{25}$

$\frac{1}{5}$

Correct answer:

$-\frac{24}{25}$

Explanation:

The double-angle identity for sine is written as

$\sin 2\theta= 2\sin\theta \cos\theta$

and we know that

$\cos\theta = \frac{4}{5}$

Using $x^{2}+y^{2}=r^{2}$ , we see that $(4)^{2} + (y)^{2} = (5)^{2}$ , which gives us

$y=\pm3$

Since we know $\theta$ is between $\frac{3\pi}{2}$ and $2\pi$ , sin $\theta$ is negative, so y= -3 . Thus,

$\sin \theta = -\frac{3}{5}$ .

Finally, substituting into our double-angle identity, we get

$\sin 2\theta = 2 \sin\theta \cos\theta = 2 \cdot\bigg(-\frac{3}{5}\bigg)\cdot\frac{4}{5} = -\frac{24}{25}$

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Example Question #1 : Identities Of Halved Angles

Find the exact value of $\sin 15^{\circ}$ using an appropriate half-angle identity.

Possible Answers:

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

$\frac{1-\sqrt{3}}{4}$

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

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

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

Correct answer:

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

Explanation:

The half-angle identity for sine is:

$\sin \bigg(\frac{x}{2}\bigg) = \pm\sqrt\frac{1-\cos x}{2}$

If our half-angle is $15^{\circ}$ , then our full angle is $30^{\circ}$ . Thus,

$\sin 15^{\circ}= \pm\sqrt\frac{1-\cos30^{\circ}}{2}$

The exact value of $\cos 30^{\circ}$ is expressed as $\frac{\sqrt{3}}{2}$ , so we have

$\sin 15^{\circ} = \pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$

Simplify under the outer radical and we get

$\sin 15^{\circ} = \pm\sqrt\frac{2-\sqrt{3}}{4}$

Now simplify the denominator and get

$\sin 15^{\circ} = \pm\frac{\sqrt{2-\sqrt{3}}}{2}$

Since $15^{\circ}$ is in the first quadrant, we know sin is positive. So,

$\sin 15^{\circ} = \frac{\sqrt{2-\sqrt{3}}}{2}$

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Example Question #3 : Identities Of Halved Angles

Which of the following best represents $2cos^2(2\theta)$ ?

Possible Answers:

$1-cos(2\theta)$

$1+cos(2\theta)$

$1-sin(4\theta)$

$1+cos(4\theta)$

$1-cos(4\theta)$

Correct answer:

$1+cos(4\theta)$

Explanation:

Write the half angle identity for cosine.

$cos^2(\theta) =\frac{1+cos(2\theta)}{2}$

Replace theta with two theta.

$\theta=2\theta$

Therefore:

$2cos^2(2\theta)= 2\left[\frac{1+cos(2 \times 2\theta)}{2}\right] = 1+cos(4\theta)$

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Example Question #4 : Identities Of Halved Angles

What is the amplitude of 8cos^2(x) ?

Possible Answers:

Correct answer:

Explanation:

The key here is to use the half-angle identity for cos^2(x) to convert it and make it much easier to work with.

$acos^2(x) = \frac{a}{2}[1 + cos(x)]$

In this case, a = 8 , so therefore...

$8cos^2(x) = \frac{8}{2}[1 + cos(x)] = 4 + 4cos(x)$

Consequently, 8cos^2(x) has an amplitude of .

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Example Question #5 : Identities Of Halved Angles

If $cos(45^{\circ})=\frac{\sqrt{2}}{2}$ , then calculate $cos(22.5^{\circ})$ .

Possible Answers:

$\frac{\sqrt{3}}{4}$

$\frac{1}{2}\sqrt{1-\sqrt{2}}$

$\frac{1}{2}\sqrt{2+\sqrt{2}}$

$\frac{1}{2}\sqrt{1+\sqrt{2}}$

$\frac{1}{2}\sqrt{2-\sqrt{2}}$

Correct answer:

$\frac{1}{2}\sqrt{2+\sqrt{2}}$

Explanation:

Because $22.5^{\circ}=\frac{45^{\circ}}{2}$ , we can use the half-angle formula for cosines to determine $cos(22.5^{\circ})$ .

In general,

$cos(\frac{\Theta}{2})= \sqrt{\frac{1}{2}(1+cos\Theta)}$

for $0^{\circ}\leq\Theta<360^{\circ}$ .

For this problem,

$cos(22.5^{\circ})= \sqrt{\frac{1}{2}(1+cos(45^{\circ}))}$

$= \sqrt{\frac{1}{2}(1+\frac{\sqrt{2}}{2})}$

$=\sqrt{\frac{1}{2}(\frac{2+\sqrt{2}}{2})}$

$=\sqrt{\frac{1}{4}(2+\sqrt{2})}$

$=\frac{1}{2}\sqrt{(2+\sqrt{2})}$

Hence,

$cos(22.5^{\circ})=\frac{1}{2}\sqrt{2+\sqrt{2}}$

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Example Question #6 : Identities Of Halved Angles

What is $\cos(\frac{5\pi}{12})$ ?

Possible Answers:

$\frac{1}{2}$

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

$\frac{\sqrt{4-\sqrt{6}}}{2}$

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

Correct answer:

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

Explanation:

Let $x=\frac{5\pi}{6}$ ; then

$\cos(\frac{5\pi}{12})=\cos(\frac{x}{2})$ .

We'll use the half-angle formula to evaluate this expression.

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

Now we'll substitute $\frac{5\pi}{6}$ for .

$\begin{align*} \cos(\frac{5\pi}{12})=\cos(\frac{\frac{5\pi}{6}}{2})&=\pm\sqrt{\frac{1+\cos(\frac{5\pi}{6})}{2}}\\ &=\pm\sqrt{\frac{1+\frac{-\sqrt{3}}{2}}{2}}\\ &=\pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\\ &=\pm\sqrt{\frac{2-\sqrt{3}}{2}}\\ &=\pm\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right) \end{align*}$

$\frac{5\pi}{6}$ is in the first quadrant, so $\cos(\frac{5\pi}{6})$ is positive. So

$\cos(\frac{5\pi}{12})=\frac{\sqrt{2-\sqrt{3}}}{2}$ .

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Example Question #51 : Trigonometry

What is $\tan(15)$ , given that $\sin(30)$ and $\cos(30)$ are well defined values?

Possible Answers:

$\frac{1}{2}$

$\frac{\frac{1}{2}}{1+\frac{\sqrt3}{2}}$

$\frac{\sqrt3}{2}$

$\frac{\frac{\sqrt3}{2}}{\frac{3}{2}}$

Correct answer:

$\frac{\frac{1}{2}}{1+\frac{\sqrt3}{2}}$

Explanation:

Using the half angle formula for tangent,

$\tan(\frac{\theta}{2})=\frac{\sin(\theta)}{(1+\cos(\theta))}$ ,

we plug in 30 for $\theta$ .

We also know from the unit circle that $\sin(30)$ is 1/2 and $\cos(30)$ is $\sqrt3/2$ .

Plug all values into the equation, and you will get the correct answer.

$\\ \tan(\frac{30}{2})=\frac{\sin(30)}{(1+\cos(30))} \\ \\ \\ \tan(15)=\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}$

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Example Question #1 : Identities Of Squared Trigonometric Functions

Using trigonometric identities prove whether the following is valid:

$\sin^{2}A + \cot^{2}A = \frac{2\sin^{4}A +2\cos^{2}A}{1 - \cos 2A}$

Possible Answers:

False

Only in the range of: $0 < A < \frac{\pi}{2}$

Only in the range of: $0 < A < 2\pi$

True

Uncertain

Correct answer:

True

Explanation:

We can work with either side of the equation as we choose. We work with the right hand side of the equation since there is an obvious double angle here. We can factor the numerator to receive the following:

$\sin^{2}A + \cot^{2}A = \frac{2\left(\sin^{4}A + \cos^{2}A\right)}{1-\cos 2A}$

Next we note the power reducing formula for sine so we can extract the necessary components as follows:

$\sin^{2}A + \cot^{2}A = \frac{2}{1-\cos 2A}\left(\sin^{4}A + \cos^{2}A\right)$

The power reducing formula must be inverted giving:

$\sin^{2}A + \cot^{2}A = \frac{1}{\sin^{2}A}\left(\sin^{4}A + \cos^{2}A\right)$

Now we can distribute and reduce:

$\sin^{2}A + \cot^{2}A = \sin^{2}A + \frac{\cos^{2}A}{\sin^{2}A}$

Finally recalling the basic identity for the cotangent:

$\sin^{2}A + \cot^{2}A = \sin^{2}A + \cot^{2}A$

This proves the equivalence.

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Example Question #1 : Trigonometric Identities

Use the power reducing formulas for trigonometric functions to reduce and simplify the following equation:

$\sin^{2}A \cos^{2} A$

Possible Answers:

$\frac{3 + \cos 4A}{8}$

$1 + \cos^{2}A$

$1 + \cos 2A$

$\frac{1 + 2\cos A}{2}$

$\frac{1 + \cos^{2}2A}{2}$

Correct answer:

$\frac{3 + \cos 4A}{8}$

Explanation:

The power reducing formulas for both sine and cosine differ in only the operation in the numerator. Applying the power reducing formulas here we get:

$\frac{1-\cos 2A}{2}\times \frac{1 + \cos 2A}{2}$

Multiplying the binomials in the numerator and multiplying the denominators:
$\frac{1-\cos 2A + \cos 2A - \cos^{2} 2A}{4}$

Reducing the numerator:

$\frac{1+\cos^{2}2A}{4}$

We again use the power reducing formula for cosine as follows:

$\frac{1 + \frac{1 + \cos 4A}{2}}{4}$

Combining the numerator by determining a common denominator:

$\frac{\frac{3 + \cos 4A}{2}}{4}$

Now simply reducing the double fraction:

$\frac{3 + \cos 4A}{8}$

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Example Question #1 : Identities Of Squared Trigonometric Functions

Using trigonometric identities, determine whether the following is valid:

$\sin^{2} A \cos^{2} A = \frac{1-\cos^{4}A-2\cos^{2}A\sin^{2}A + \sin^{4}A}{4}$

Possible Answers:

True

Only valid in the range of: $0 < A < \frac{\pi}{2}$

Uncertain

False

Only valid in the range of: $0 < A < 2\pi$

Correct answer:

False

Explanation:

In this case we choose to work with the side that appears to be simpler, the left hand side. We begin by using the power reducing formulas:

$\frac{1-\cos 2A}{2}\frac{1+\cos 2A}{2} = \frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

Next we perform the multiplication on the numerator:

$\frac{1-\cos^{2}2A}{4} = \frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

The next step we take is to remove the double angle, since there is no double angle in the alleged solution:

$\frac{1 - \left(\cos^{2}A -\sin^{2}A\right )^{2}}{4}=\frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

Finally we multiply the binomials in the numerator on the left hand side to determine if the equivalence holds:

$\frac{1-\cos^{4}A +2\cos^{2}A\sin^{2}A-\sin^{4}A}{4} = \frac{1-\cos^{4}A -2\cos^{2}A\sin^{2}A+\sin^{4}A}{4}$

We see that the equivalence does not hold.

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Trigonometry : Trigonometric Identities

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #51 : Trigonometric Identities

Example Question #1 : Identities Of Halved Angles

Example Question #3 : Identities Of Halved Angles

Example Question #4 : Identities Of Halved Angles

Example Question #5 : Identities Of Halved Angles

Example Question #6 : Identities Of Halved Angles

Example Question #51 : Trigonometry

Example Question #1 : Identities Of Squared Trigonometric Functions

Example Question #1 : Trigonometric Identities

Example Question #1 : Identities Of Squared Trigonometric Functions

All Trigonometry Resources

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