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

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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Question

Simplify

$\frac{\sin^{3}{\left (x \right )} + \cos{\left (x \right )}}{2 \cos^{2}{\left (x \right )} - 1}$

Answer

The first step to simplifying is to remember an important trig identity.

$\sin^{2}{\left (x \right )} + \cos^{2}{\left (x \right )} = 1$

If we rewrite it to look like the denominator, it is.

$\sin^{2}{\left (x \right )} = \cos^{2}{\left (x \right )} - 1$

Now we can substitute this in the denominator.

$\frac{\sin^{3}{\left (x \right )} + \cos{\left (x \right )}}{2 \sin^{2}{\left (x \right )}}$

Now write each term separately.

$\frac{1}{2} \sin{\left (x \right )} + \frac{\cos{\left (x \right )}}{2 \sin^{2}{\left (x \right )}}$

Remember the following identities.

$\\csc{\left (x \right )} = \frac{1}{\sin{\left (x \right )}} \\\sec{\left (x \right )} = \frac{1}{\cos{\left (x \right )}} \\\tan{\left (x \right )} = \frac{\sin{\left (x \right )}}{\cos{\left (x \right )}} \\\cot{\left (x \right )} = \frac{\cos{\left (x \right )}}{\sin{\left (x \right )}}$

Now simplify, and combine each term.

$\\frac{1}{2} \sin{\left (x \right )} + \frac{1}{2} \cos{\left (x \right )} \csc^{2}{\left (x \right )} \\\frac{1}{2} \left(\sin{\left (x \right )} + \cos{\left (x \right )} \csc^{2}{\left (x \right )}\right)$

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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

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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