Trigonometric Identities - Trigonometry

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Question

Which of the following identities is incorrect?

Answer

The true identity is $\small \cos(-x)=\cos(x)$ because cosine is an even function.

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Question

Which of the following trigonometric identities is INCORRECT?

Answer

Cosine and sine are not reciprocal functions.

$sin (x) = \frac{1}{csc (x)}$ and $cos (x) = \frac{1}{sec (x)}$

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Question

Using the trigonometric identities prove whether the following is valid:

$\frac{\csc^{2}A}{\sec^{2}A} = \cot^{2}A$

Answer

We begin with the left hand side of the equation and utilize basic trigonometric identities, beginning with converting the inverse functions to their corresponding base functions:

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

Next we rewrite the fractional division in order to simplify the equation:

$\frac{1}{\sin^{2}A}\div\frac{1}{\cos^{2}A} = \cot^{2}A$

In fractional division we multiply by the reciprocal as follows:

$\frac{1}{\sin^{2}A}\times \cos^{2}A = \cot^{2}A$

If we reduce the fraction using basic identities we see that the equivalence is proven:

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

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

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Question

Which of the following is the best answer for $cos^2(\theta)$ ?

Answer

Write the Pythagorean identity.

$sin^2(\theta)+cos^2(\theta)=1$

Substract $sin^2(\theta)$ from both sides.

$cos^2(\theta)=1-sin^2(\theta)$

The other answers are incorrect.

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Question

State $\tan x$ in terms of sine and cosine.

Answer

The definition of tangent is sine divided by cosine.

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

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Question

Simplify.

$\sin(-x)\cos(-x)\tan(-x)$

Answer

Using these basic identities:

$\sin(-x)=-\sin x$

$\cos(-x)=\cos(x)$

$\tan(-x)=-\tan x$

we find the original expression to be

$(-\sin x)(\cos x)(-\tan x)$

which simplifies to

$\sin x\cos x\tan x$ .

Further simplifying:

$\sin x\cos x\frac{\sin x}{\cos x}$

The cosines cancel, giving us

$\sin ^2 x$

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Question

Express $\frac{1}{\sec x \csc x \tan x}$ in terms of only sines and cosines.

Answer

The correct answer is $\cos^2 x$ . Begin by substituting $\frac{1}{\sec x}=\cos x$ , $\frac{1}{\csc x}=\sin x$ , and $\frac{1}{\tan x}=\frac{\cos x}{\sin x}$ . This gives us:

$\frac{1}{\sec x\csc x\tan x}=\frac{\cos x\sin x\cos x}{\sin x}=\cos^2 x$ .

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Question

Express $\sec x\tan x+\frac{\cot x}{\csc x}$ in terms of only sines and cosines.

Answer

To solve this problem, use the identities $\sec x =\frac{1}{\cos x}$ , $\tan x=\frac{\sin x}{\cos x}$ , $\cot x =\frac{\cos x}{\sin x}$ , and $\frac{1}{\csc x}=\sin x$ . Then we get

$\sec x\tan x+\frac{\cot x}{\csc x}$

$=\frac{1}{\cos x}\cdot \frac{\sin x}{\cos x}+\sin x\cdot \frac{\cos x}{\sin x}$

$=\frac{\sin x}{\cos^2 x}+\cos x$

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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

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

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

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

Using trigonometric identities determine whether the following is valid:

$\tan2A = \frac{2\sin \left(-A\right ) \cos \left(-A\right )}{\cos^{2}\left(-A\right ) + \sin^{2}\left(-A\right)}$

Answer

We can choose either side to work with to attempt to obtain the equivalency. Here we will work with the right side as it is the more complex. First, we want to eliminate the negative angles using the appropriate relations. Sine is odd and therefore, the negative sign comes out front. Cosine is even which is interpreted by dropping the negative out of the equation:

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

The squaring of the sine in the denominator makes the sine term positive, i.e.

$\sin^{2}\left(-A \right ) = \left(\sin-A \right )\times\left(\sin -A \right ) = -\sin A \times -\sin A = \sin^{2}A$

The numerator is the double angle formula for sine:

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

The denominator is recognized to be the pythagorean theorem as it applies to trigonometry:

$\tan 2A = \frac{-2\sin 2A}{1}$

The final reduced equation is:

$\tan 2A = -2\sin 2A$

Thus proving that the equivalence is false.

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Question

You can derive the formula $1+\tan^2\theta=\sec^2\theta$ by dividing the formula $\sin^2\theta+\cos^2\theta=1$ by which of the following functions?

Answer

The correct answer is $\cos^2\theta$ . Rather than memorizing all three Pythagorean Relationships, you can memorize only $\sin^2\theta+\cos^2\theta=1$ , then simply divide all terms by $\cos^2\theta$ to get the formula that relates $\tan^2\theta$ and $\sec^2\theta$ . Alternatively, you can divide all terms of $\sin^2\theta+\cos^2\theta=1$ by $\sin^2\theta$ to get the formula that relates $\cot^2\theta$ and $\csc^2\theta$ . The former is demonstrated below.

$\sin^2\theta+\cos^2\theta=1$

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

$\tan^2\theta+1=\sec^2\theta$

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Question

You can derive the formula $1+\cot^2\theta=\csc^2\theta$ by dividing the formula $\sin^2\theta+\cos^2\theta=1$ by which of the following functions?

Answer

The correct answer is $\sin^2\theta$ . Rather than memorizing all three Pythagorean Relationships, you can memorize only $\sin^2\theta+\cos^2\theta=1$ , then simply divide all terms by $\sin^2\theta$ to get the formula that relates $\cot^2\theta$ and $\csc^2\theta$ . Alternatively, you can divide all terms of $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ to get the formula that relates $\tan^2\theta$ and $\sec^2\theta$ . The former is demonstrated below.

$\sin^2\theta+\cos^2\theta=1$

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

$1+\cot^2\theta=\csc^2\theta$

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Question

One popular way to simplify trigonometric expressions is to put the entire expression in terms of only and functions.

Answer

Getting every term in an expression in terms of sine and cosine functions is a popular way to verify trigonometric identities or complete trigonometry proofs. These two trig functions are more commonly used over their counterparts secant, cosecant, tangent, and cotangent. Moreover, getting all terms of an expression in strictly sine and cosine may help you to spot and then substitute $\sin^2\theta+\cos^2\theta=1$ , or it may help you spot other functions that can be reduced or simplified. Other general techniques to aid in verifying trigonometric identities are:

Know the eight basic relationship and recognize alternative forms of each. These are: $\csc\theta=\frac{1}{\sin\theta}$ , $\sec\theta=\frac{1}{\cos\theta}$ , $\cot\theta=\frac{1}{\tan\theta}$ , $\tan\theta=\frac{\sin\theta}{\cos\theta}$ , $\cot\theta=\frac{\cos\theta}{\sin\theta}$ , $\sin^2\theta+\cos^2\theta=1$ , $1+\tan^2\theta=\sec^2\theta$ , and $1+\cot^2\theta=\csc^2\theta$ .

Understand how to add and subtract fractions, reduce fractions, and transform fractions into equivalent fractions

Know how to factor, and know special product techniques (i.e. difference of squares)

Only work on one side of the equation at a time

Choose the side of the equation that looks more complicated and attempt to transform it into the other side of the equation

Alternatively, you may transform each side of the equation into the same form. If doing this, you may need to do scratch work, then go back and nicely organize the two transformed sides into one clean looking verification.

Avoid substitutions that introduce radicals.

Multiply the numerator and denominator of a fraction by the conjugate of either.

Simplify a square root of a fraction by using conjugates to transform it into the quotient of a perfect squares.

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Question

Fill in the four blanks, naming the appropriate trigonometric functions, and fill in the two question marks to complete the proofs showing that $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\cot\theta=\frac{\cos\theta}{\sin\theta}$ .

For any angle $\theta$ :

$(\theta)=\frac{y}{r}$ , $\theta=\frac{x}{r}$ , $\theta=\frac{y}{x}$ , and $\theta=\frac{x}{y}$ where is any point on the terminal side of $\theta$ and is units from the origin.

Then, $\tan\theta=?=\frac{\frac{y}{r}}{\frac{x}{r}}=\frac{\sin\theta}{\cos\theta}$ and $\cot\theta=?=\frac{\frac{x}{r}}{\frac{y}{r}}=\frac{\cos\theta}{\sin\theta}$

Answer

This question is asking us to fill in the first four blanks by adding the appropriate trigonometric function that satisfies each. By defining an angle $\theta$ on the coordinate plane, and labelling the distances x, y, and r, we can ask ourselves which trig functions relate each of the following sides together. Many students use the acronym SOHCAHTOA to remember these. Using the diagram below, we can see that $\sin\theta=\frac{opp}{hyp}=\frac{y}{r}$ , $\cos\theta=\frac{adj}{hyp}=\frac{x}{r}$ , and $\tan\theta=\frac{opp}{adj}=\frac{y}{x}$ . Knowing each of these three answers, the fourth answer (based on the available answer choices) must be $\cot\theta=\frac{x}{y}$ .

Next, we need to fill in the two question marks. To do this, simply substitute the info from the first part of the proof to get that $\tan\theta=\frac{y}{x}=\frac{\frac{y}{r}}{\frac{x}{r}}$ and $\cot\theta=\frac{x}{y}=\frac{\frac{x}{r}}{\frac{y}{r}}$ .

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Question

Which of the following might be an effective first step in verifying the following identity? Assume that you are going to attempt to convert the left hand side into the right hand side.

$\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}=\frac{1}{\sec x+\tan x}$

Answer

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods.

The correct answer to how to best begin this verification is "Simplify a square root of a fraction by using conjugates to transform it into the quotient of perfect squares." This technique will be effective because while the left side of the equation has a square root and the right hand side does not, eliminating that square root will be crucial to verifying the identity. The verification of this identity is demonstrated below:

$\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}=\sqrt{\frac{(\sec x-\tan x)\cdot (\sec x+\tan x)}{(\sec x+\tan x)\cdot (\sec x+\tan x)}}=\sqrt{\frac{\sec^2 x-\tan^2 x}{(\sec x+\tan x)^2}}$ $=\sqrt{\frac{1}{(\sec x+\tan x)^2}}=\frac{1}{\sec x+\tan x}$

Let's also explore why the incorrect answers are incorrect.

Firstly, while getting everything in terms of sines and cosines is a popular strategy in verifying trig identities, it will actually introduce more fractions, while failing to get rid of the square root function. Next, this problem does not lend itself to substituting a Pythagorean Identity. Pythagorean Identities are most helpful when you can use them to replace something like $\sin^2 x+\cos^2 x$ with . They can also be helpful if you want to replace something like $\sin^2 x$ with $1-\cos^2 x$ (while this seems more complicated to replace one term with two terms, if all of the rest of the terms in the expression were all cosines, this could be helpful.) Next, consider factoring. While often a helpful strategy, this method lends itself best to when you have polynomial terms such as $(\cos^2x-4)$ , which could be factored into $(\cos x+2)(\cos x-2)$ . This problem does not have this type of expression, so this isn't helpful to us. Finally, reducing fractions can be a great strategy when there are fractions to reduce. Do not fall into the trap of thinking that you are able to cancel the $\sec x$ or $\tan x$ expressions on the left hand side of the original equations. If you wanted to split this up, the only way to properly do it would be to get $\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}=\sqrt{\frac{\sec x}{\sec x+\tan x}-\frac{\tan x}{\sec x+\tan x}}$ . Notice how large this expression is, and notice that it is getting more complicated rather than simpler and closer to $\frac{1}{\sec x+\tan x}$ . Often times, when trying to verify trig identities, you may start with one approach and find it not fruitful, and need to go back to the start and try again. Always remember that there may be another way to approach the problem; be flexible and be willing to experiment and try more than one approach!

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