Trigonometric Identities - Trigonometry

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Question

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

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

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Answer

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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Question

Using trigonometric identities prove whether the following is valid:

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

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Answer

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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Question

Using trigonometric identities, determine whether the following is valid:

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

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Answer

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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Question

Simplify the expression:

$\csc\theta(\sin\theta+\cos\theta \cot\theta)$

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Answer

The first step in solving this equation is to distribute :

$\csc\theta(\sin\theta +\cos\theta \cot\theta ) = \csc \theta\sin\theta+ \csc\theta\cos\theta\cot\theta$

At this point, simplify using known Pythagorean identities. The left quantity simplifies such:

$\csc\theta\sin\theta = \frac{\sin\theta }{\sin\theta } = 1$

and the right quantity simplifies such:

$\csc\theta\cos\theta\cot\theta = (\frac{1}{\sin\theta })(\cos\theta )(\frac{\cos\theta }{\sin\theta })=\frac{\cos^{2}\theta }{\sin^{2}\theta }=\cot^{2}\theta$

Thus, we end up with:

$1+\cot^2\theta$ ,

which our Pythagorean identity tells us is equivalent to $\csc^{2}\theta$ .

Thus,

$\csc\theta(\sin\theta+\cos\theta\cot\theta) = \csc^{2}\theta$

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Question

Find the exact value of the expression:

$\sin 60^{\circ}\cos 30^{\circ} + \cos 60^{\circ} \sin 30^{\circ}$

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Answer

There are two ways to solve this problem. If one recognizes the identity

$\sin (a+b)= \sin a\cos b + \cos a\sin b$ ,

the answer is as simple as:

$\sin 60^{\circ}\cos 30^{\circ} + \cos 60^{\circ}\sin 30^{\circ} = \sin (60^{\circ}+30^{\circ}) = \sin 90^{\circ} = 1$

If one misses the identity, or wishes to be more thorough, you can simplify:

$\sin 60^{\circ}\cos 30^{\circ} + \cos 60^{\circ}\sin 30^{\circ}$

$=\bigg (\bigg(\frac{\sqrt{3}}{2}\bigg)\cdot\bigg(\frac{\sqrt{3}}{2}\bigg)\bigg)+\bigg(\bigg(\frac{1}{2}\bigg)\cdot\bigg(\frac{1}{2}\bigg)\bigg)=\frac{3}{4} + \frac{1}{4}=1$

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Question

Simplify using identities:

$\frac{-\cos x+\sec x}{\tan x}$

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Answer

First we expand the inverse identities into fractional form:

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

Invert the bottom fraction and distribute into the top, keeping track of the negative:

$\frac{-cos x+\frac{1}{\cos x}}{\frac{\sin x}{\cos x}} = \bigg(\frac{1-\cos^{2}x}{\cos x}\bigg)\bigg(\frac{\cos x}{\sin x}\bigg)$

Using the Pythagorean identity $1-\cos^{2}x = \sin^{2}x$ , our equation becomes:

$\bigg(\frac{\sin^{2}x}{\cos x}\bigg)\bigg(\frac{\cos x}{\sin x}\bigg)$

At this point, cross-cancel to obtain $\sin x$ .

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Question

Determine the value of: $(1+cot^2(\theta))(1-cos^2(\theta))$

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Answer

In order to solve $(1+cot^2(\theta))(1-cos^2(\theta))$ , first identify all the recognizable identities.

The identities that can be used for this scenario are:

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

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

The first identity can also be manipulated.

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

Replace $(1+cot^2(\theta))(1-cos^2(\theta))$ with the correct identities and simplify.

$csc^2(\theta)\times sin^2(\theta)= \frac{1}{sin^2(\theta)}\times sin^2(\theta) =1$

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Question

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

Which of the following is equivalent to the function above.

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

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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

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

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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?

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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?

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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:

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

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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 $cos^2(\theta)$ ?

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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.

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Answer

The definition of tangent is sine divided by cosine.

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

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Question

Simplify.

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

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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.

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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.

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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 ?

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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?

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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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