Trigonometric Identities - Trigonometry

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Question

$f(x)=\sin(x)$

Which of the following is equivalent to the function above.

Answer

The cosine graph at zero has a peak at 1.

The first peak in the sine curve is at π/2, so you adjust the cosine with a -π/2 and that gives the answer $\cos(x-\frac{\pi}{2})$

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Question

Which of those below is equivalent to ?

Answer

Think of a right triangle. The two non-right angles are complementary since the angles in a triangle add up to 180. As a result, when we apply the definitions of sine and cosine, we get that the sine of one of these angles is equivalent to the cosine of the other. As a result, the sine of one angle is equal to the cosine of its complement and vice versa.

Therefore, when we look at the angles of a right triangle that includes a 60 degree angle we get

.

Thus the

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Question

Given that $\sin 60^{\circ}= z$ , which of the following must also be true?

Answer

The supplementary angle identity states that in the positive quadrants, if two angles are supplementary (add to 180 degrees), they must have the same sine. In other words,

if $0^{\circ}<\theta<180^{\circ}$ , then $\sin A = \sin(180-A)$

Using this, we can see that

$\sin 60^{\circ} = \sin (180^{\circ}-60^{\circ}) = \sin 120^{\circ}$

Thus, if $\sin 60^{\circ} = z$ , then $\sin 120^{\circ} = z$ also.

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Question

Which of the following is equivalent to ?

Answer

Recall that the cosine and sine of complementary angles are equal.

Thus, we are looking for the complement of 70, which gives us 20.

When we take the cosine of this, we will get an equivalent statement.

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Question

Simplify the following expression:

$\frac{cos(90-x)}{tan(x))}$

Answer

Simplifying this expression relies on an understanding of the cofunction identities. The cofunction identities hinge on the fact that the value of a trig function of a particular angle is equal to the value of the cofunction of the the complement of that angle. In other words,

That means that returning to our initial expression, we can do some substiution.

$\frac{cos(90-x)}{tan(x)}=\frac{sin(x)}{tan(x)}$

We then turn to another identity, namely the fact that tangent is just the quotient of sine and cosine.

$tan(x)=\frac{sin(x)}{cos(x)}$

We can substitute again

$\frac{sin(x)}{tan(x)}=\frac{sin(x)}{\frac{sin(x)}{cos(x)}}$

Yet dividing by a fraction is the same as multipying by the reciprocal.

$\frac{sin(x)}{\frac{sin(x)}{cos(x)}}=sin(x)\cdot\frac{cos(x)}{sin(x)}=cos(x)$

With some cancellation, we have arrived at our answer.

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Question

Which one is equal to $\small \sin 44$

Answer

Complementary angles are equal to one's $\cos(x)$ to others $\sin(90-x)$

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Question

Find $sin (75 ^{\circ})$ without using a calculator.

Answer

To answer this question, use the sum formula for sine:

$sin (75 ^{\circ}) = sin (45 ^{\circ} + 30^{\circ}) =$

$sin(45^{\circ})cos(30^{\circ}) + cos (45^{\circ})sin(30^{\circ}) =$

$\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$

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Question

Simplify the following expression using trigonometric identities:

$\sin^{3}A\cos B + \sin^{2}A\cos A\sin B + \sin A\cos B\cos^{2}A + \cos^{3}A\sin B$

Answer

In order to simplify the given equation we should first try to determine if the Pythagorean Theorem as applicable to trigonomety can be utilized. We do this first due to the higher degree of the functions involved. We can notice that if we group the higher order sine and the higher order cosine, that we can in fact pull out some common terms:

$\sin^{2}A\left(\sin A \cos B + \cos A \sin B\right) + \cos^{2} A \left(\sin A \cos B + \cos A \sin B\right )$

Now we notice that we can further group the terms:

$\left(\sin^{2}A + \cos^{2} A\right)\left(\sin A \cos B + \cos A \sin B\right)$

The first term in the previous equation is in fact the Pythagorean Theorem as applied to trigonometry and the second term is the sum of two angles with respect to the sine function:

$\left(1\right)\left(\sin \left(A + B\right ) \right )$

This reduced simply to the sum function for sine:

$\sin(A + B)$

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Question

Without the aide of a calculator, compute the exact value of $\sin\left(\frac{7\pi}{12} \right )$

Answer

In order to determine an exact value we want to split up the angle into angles which we can determine the exact value; these are multiples of $\frac{\pi}{6}$ , $\frac{\pi}{4}$ , $\frac{\pi}{3}$ , $\frac{\pi}{2}$ , and $\pi$ . Since the given angle has a 12 in the denominator we must have two fractions which can be found to have a common denominator of 12, the pairs are 3 and 4 or 2 and 6. We can write an algebraic equation which must hold to assist us in determining these values:

$\frac{A\pi}{3} \pm \frac{B\pi}{4} = \frac{7\pi}{12}$

Where and are integers. If we work to get a common denominator of 12:

$\frac{4A\pi}{12} \pm \frac{3B\pi}{12} = \frac{7\pi}{12}$

We can now reduce the equation to obtain a simpler form:

$4A \pm 3B = 7$

Using this equation it is simple to see that and and the operation is addition. Therefore rewriting the equation:

$\sin\left(\frac{7\pi}{12} \right ) = \sin\left(\frac{\pi}{3} + \frac{\pi}{4} \right )$

Where we have simply reduced the fractions to their lowest terms. Now that we have the sum of two known angles we can simply continue and write out the appropriate formula:

$\sin\left(\frac{\pi}{3} +\frac{\pi}{4}\right ) = \sin\left(\frac{\pi}{3} \right )\cos\left(\frac{\pi}{4} \right ) + \cos\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{4} \right )$

We know the values of the corresponding angles and we must note the quadrant the angle is in, here we are in the first quadrant for both angles which means that all values are positive.

$\sin\left(\frac{\pi}{3} +\frac{\pi}{4}\right ) = \frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2} + \frac{1}{2}\frac{\sqrt{2}}{2}$

This reduces to:
$\sin\left(\frac{\pi}{3} +\frac{\pi}{4}\right ) = \frac{\sqrt{6} + \sqrt{2}}{4}$

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Question

Without the aide of a calculator, compute the exact value of $\sin\left(\frac{\pi}{12} \right )$

Answer

In order to determine an exact value we want to split up the angle into angles which we can determine the exact value; these are multiples of $\frac{\pi}{6}$ , $\frac{\pi}{4}$ , $\frac{\pi}{3}$ , $\frac{\pi}{2}$ , and $\pi$ . Since the given angle has a 12 in the denominator we must have two fractions which can be found to have a common denominator of 12, the pairs are 3 and 4 or 2 and 6. We can write an algebraic equation which must hold to assist us in determining these values:

$\frac{A\pi}{3} \pm \frac{B\pi}{4} = \frac{\pi}{12}$

Where and are integers. If we work to get a common denominator of 12:

$\frac{4A\pi}{12} \pm \frac{3B\pi}{12} = \frac{\pi}{12}$

We can now reduce the equation to obtain a simpler form:

$4A \pm 3B = 1$

Using this equation it is simple to see that and and the operation is subtraction. Therefore rewriting the equation:

$\sin\left(\frac{\pi}{12} \right ) = \sin\left(\frac{\pi}{3} - \frac{\pi}{4} \right )$

Where we have simply reduced the fractions to their lowest terms. Now that we have the sum of two known angles we can simply continue and write out the appropriate formula:

$\sin\left(\frac{\pi}{3} -\frac{\pi}{4}\right ) = \sin\left(\frac{\pi}{3} \right )\cos\left(\frac{\pi}{4} \right ) - \cos\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{4} \right )$

We know the values of the corresponding angles and we must note the quadrant the angle is in, here we are in the first quadrant for both angles which means that all values are positive.

$\sin\left(\frac{\pi}{3} -\frac{\pi}{4}\right ) = \frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2} - \frac{1}{2}\frac{\sqrt{2}}{2}$

This reduces to:
$\sin\left(\frac{\pi}{3} -\frac{\pi}{4}\right ) = \frac{\sqrt{6} - \sqrt{2}}{4}$

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Question

Without the aide of a calculator, compute the exact value of $\cos\left(-\frac{5\pi}{12} \right )$

Answer

In order to determine an exact value we want to split up the angle into angles which we can determine the exact value; these are multiples of $\frac{\pi}{6}$ , $\frac{\pi}{4}$ , $\frac{\pi}{3}$ , $\frac{\pi}{2}$ , and $\pi$ . Since the given angle has a 12 in the denominator we must have two fractions which can be found to have a common denominator of 12, the pairs are 3 and 4 or 2 and 6. We can write an algebraic equation which must hold to assist us in determining these values:

$\frac{A\pi}{3} \pm \frac{B\pi}{4} = -\frac{5\pi}{12}$

Where and are integers. If we work to get a common denominator of 12:

$\frac{4A\pi}{12} \pm \frac{3B\pi}{12} = -\frac{5\pi}{12}$

We can now reduce the equation to obtain a simpler form:

$4A \pm 3B = -5$

Using this equation we actually need to flip the fractions as follows:

$3B \pm 4A = -5$

Now we can get the right values of; and and the operation is subtraction. Therefore rewriting the equation:

$\cos\left(\frac{\pi}{12} \right ) = \cos\left(\frac{\pi}{4} - \frac{2\pi}{3} \right )$

Where we have simply reduced the fractions to their lowest terms. Now that we have the sum of two known angles we can simply continue and write out the appropriate formula:

$\cos\left(\frac{\pi}{4} - \frac{2\pi}{3} \right ) = \cos\left(\frac{\pi}{4} \right )\cos\left(\frac{2\pi}{3}\right) + \sin\left(\frac{\pi}{4} \right )\sin\left(\frac{2\pi}{3}\right)$

We know the values of the corresponding angles and we must note the quadrant the angle is in, here we are in the first quadrant for angle A which means that all values are positive and the second angle, B, we are in the second quadrant so sine is positive and all other functions are negative.

$\cos\left(\frac{\pi}{4} - \frac{2\pi}{3} \right ) = \frac{\sqrt{2}}{2}\left(-\frac{1}{2} \right ) + \frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

This reduces to:

$\cos\left(\frac{\pi}{4} -\frac{2\pi}{3}\right ) = \frac{-\sqrt{2} + \sqrt{6}}{4}$

Finally, we obtain the solution:

$\cos\left(\frac{\pi}{4} -\frac{2\pi}{3}\right ) = \frac{\sqrt{6}- \sqrt2}{4}$

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Question

What is the phase shift of $f(x) = sin(x)cos(\frac{\pi}{2}) - cos(x)sin(\frac{\pi}{2})$ ?

Answer

The key here is to use this sum/product identity:

$sin(u \pm v) = sin(u)cos(v) \pm cos(u)sin(v)$

In this case, and $v = \frac{\pi}{2}$ . Note as well that because subtracts, it will translate into . So using the identity, we can state as...

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

Bear in mind that for a sine function of form

...our phase shift is equal to...

$\frac{c}{b} = \frac{(\frac{\pi}{2})}{1} = \frac{\pi}{2}$

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Question

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

Answer

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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Question

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

Answer

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 . 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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Question

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

Answer

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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Question

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

Answer

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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Question

What is the amplitude of ?

Answer

The key here is to use the half-angle identity for to convert it and make it much easier to work with.

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

In this case, , so therefore...

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

Consequently, has an amplitude of .

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Question

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

Answer

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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Question

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

Answer

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 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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Question

Simplify $\frac{1-\cos^2\theta}{\sin\theta\cos\theta}$ .

Answer

Recognize that $1-\cos^2\theta$ is a reworking on $\sin^2+\cos^2=1$ , meaning that $1-\cos^2\theta=\sin^2\theta$ .

Plug that in to our given equation:

$\frac{1-\cos^2\theta}{\sin\theta\cos\theta}=\frac{\sin^2\theta}{\sin\theta\cos\theta}$

Notice that one of the $\sin\theta$ 's cancel out.

$\frac{\sin\theta}{\cos\theta}=\tan\theta$ .

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