Apply Basic and Definitional Identities - Trigonometry

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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 the following identities is incorrect?

Tap to reveal answer

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?

Tap to reveal answer

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:

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

Tap to reveal answer

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 $cos^2(\theta)$ ?

Tap to reveal answer

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.

Tap to reveal answer

Answer

The definition of tangent is sine divided by cosine.

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

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Question

Simplify.

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

Tap to reveal answer

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.

Tap to reveal answer

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.

Tap to reveal answer

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 the following identities is incorrect?

Tap to reveal answer

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?

Tap to reveal answer

Answer

Cosine and sine are not reciprocal functions.

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

← Didn't Know|Knew It →

Question

Using the trigonometric identities prove whether the following is valid:

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

Tap to reveal answer

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:

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

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

Knew It

Apply Basic and Definitional Identities - Trigonometry

Which of the following identities is incorrect?

The true identity is because cosine is an even function.

Which of the following trigonometric identities is INCORRECT?

Cosine and sine are not reciprocal functions. and

Using the trigonometric identities prove whether the following is valid:

Which of the following is the best answer for ?

Write the Pythagorean identity. Substract from both sides. The other answers are incorrect.

State in terms of sine and cosine.

The definition of tangent is sine divided by cosine.

Simplify.

Using these basic identities: we find the original expression to be which simplifies to . Further simplifying: The cosines cancel, giving us

Express in terms of only sines and cosines.

The correct answer is . Begin by substituting , , and . This gives us: .

Express in terms of only sines and cosines.

To solve this problem, use the identities , , , and . Then we get

Which of the following identities is incorrect?

The true identity is because cosine is an even function.

Which of the following trigonometric identities is INCORRECT?

Cosine and sine are not reciprocal functions. and

Using the trigonometric identities prove whether the following is valid:

Which of the following is the best answer for ?

Write the Pythagorean identity. Substract from both sides. The other answers are incorrect.

State in terms of sine and cosine.

The definition of tangent is sine divided by cosine.

Simplify.

Using these basic identities: we find the original expression to be which simplifies to . Further simplifying: The cosines cancel, giving us

Express in terms of only sines and cosines.

The correct answer is . Begin by substituting , , and . This gives us: .

Express in terms of only sines and cosines.

To solve this problem, use the identities , , , and . Then we get

Which of the following identities is incorrect?

The true identity is because cosine is an even function.

Which of the following trigonometric identities is INCORRECT?

Cosine and sine are not reciprocal functions. and

Using the trigonometric identities prove whether the following is valid:

Which of the following is the best answer for ?

Write the Pythagorean identity. Substract from both sides. The other answers are incorrect.

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

Cosine and sine are not reciprocal functions.

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

Using the trigonometric identities prove whether the following is valid:

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

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

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.

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

The definition of tangent is sine divided by cosine.

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

Simplify.

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

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

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:

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

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

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

Cosine and sine are not reciprocal functions.

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

Using the trigonometric identities prove whether the following is valid:

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

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

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.

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

The definition of tangent is sine divided by cosine.

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

Simplify.

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

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

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:

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

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

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

Cosine and sine are not reciprocal functions.

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

Using the trigonometric identities prove whether the following is valid:

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

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

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.