Trigonometric Identities

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Using trigonometric identities prove whether the following is valid:

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

True

False

Uncertain

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

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

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.

Simplify

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

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

$\frac{1}{2} \cos^{3}{\left (x \right )} \csc^{2}{\left (x \right )}$

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

$\sin^{2}{\left (x \right )} + \sec^{2}{\left (x \right )}$

$\frac{1}{2}$

Explanation

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.

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

Now write each term separately.

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

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.

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

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?

$\cos\theta$

$\sin\theta$

$\cos^2\theta$

$\sin^2\theta$

$\tan^2\theta$

Explanation

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.

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

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

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

Using trigonometric identities prove whether the following is valid:

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

True

False

Uncertain

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

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

Explanation

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

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.

Simplify

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

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

$\frac{1}{2} \cos^{3}{\left (x \right )} \csc^{2}{\left (x \right )}$

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

$\sin^{2}{\left (x \right )} + \sec^{2}{\left (x \right )}$

$\frac{1}{2}$

Explanation

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.

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

Now write each term separately.

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

Remember the following identities.

Now simplify, and combine each term.

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

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?

$\cos\theta$

$\sin\theta$

$\cos^2\theta$

$\sin^2\theta$

$\tan^2\theta$

Explanation

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

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

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

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?

$\sin\theta$

$\cos\theta$

$\cos^2\theta$

$\sin^2\theta$

$\tan^2\theta$

Explanation

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.

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

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

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

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?

$\sin\theta$

$\cos\theta$

$\cos^2\theta$

$\sin^2\theta$

$\tan^2\theta$

Explanation

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

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

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

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

sine, tangent

cosine, tangent

cotangent, tangent

cotangent, cosine

sine, cosine

Explanation

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.

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

sine, tangent

cosine, tangent

cotangent, tangent

cotangent, cosine

sine, cosine

Explanation

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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