Math / Trigonometry

QuestionEach of the following are equal to $\operatorname{Sin}(A)$ EXCEPT
A $\sin (-A)$ B $\sin (\pi-A)$ (C) $\cos \left(90^{\circ}-A\right)$
D $\frac{1}{\csc A}$
SUBMIT ANSWER

Studdy Solution

STEP 1

What is this asking? Which of these expressions is not equal to $\sin(A)$ ? Watch out! Remember your trigonometric identities and the relationships between sine and cosine!
Also, be careful with signs and angles.

STEP 2

1. Analyze option A
2. Analyze option B
3. Analyze option C
4. Analyze option D

STEP 3

We're given $\sin(-A)$ .
The sine function has a special property: it's an odd function!
This means $\sin(-A) = -\sin(A)$ .
So, option A is not equal to $\sin(A)$ .

STEP 4

We have $\sin(\pi - A)$ .
Remember the sine subtraction formula: $\sin(x - y) = \sin(x)\cos(y) - \cos(x)\sin(y)$ .

STEP 5

Let's apply this to our expression: $\sin(\pi - A) = \sin(\pi)\cos(A) - \cos(\pi)\sin(A)$

STEP 6

We know that $\sin(\pi) = \textbf{0}$ and $\cos(\pi) = \textbf{-1}$ .
Substituting these values, we get: $\sin(\pi - A) = (\textbf{0})\cos(A) - (\textbf{-1})\sin(A) = \sin(A)$ So, option B is equal to $\sin(A)$ .

STEP 7

We're looking at $\cos(90^\circ - A)$ .
This reminds us of the cofunction identity: $\cos(90^\circ - A) = \sin(A)$ .
So, option C is equal to $\sin(A)$ .

STEP 8

We have $\frac{1}{\csc(A)}$ .
Remember that $\csc(A)$ is the reciprocal of $\sin(A)$ , meaning $\csc(A) = \frac{1}{\sin(A)}$ .

STEP 9

Therefore, $\frac{1}{\csc(A)} = \frac{1}{\frac{1}{\sin(A)}} = \sin(A)$ .
So, option D is equal to $\sin(A)$ .

STEP 10

Options B, C, and D are all equal to $\sin(A)$ .
Only option A, which simplifies to $-\sin(A)$ , is not equal to $\sin(A)$ .
Thus, the answer is A.

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QuestionEach of the following are equal to $\operatorname{Sin}(A)$ EXCEPT
A $\sin (-A)$ B $\sin (\pi-A)$ (C) $\cos \left(90^{\circ}-A\right)$
D $\frac{1}{\csc A}$
SUBMIT ANSWER

Studdy Solution

STEP 1

What is this asking? Which of these expressions is not equal to $\sin(A)$ ? Watch out! Remember your trigonometric identities and the relationships between sine and cosine!
Also, be careful with signs and angles.

STEP 2

1. Analyze option A
2. Analyze option B
3. Analyze option C
4. Analyze option D

STEP 3

We're given $\sin(-A)$ .
The sine function has a special property: it's an odd function!
This means $\sin(-A) = -\sin(A)$ .
So, option A is not equal to $\sin(A)$ .

STEP 4

We have $\sin(\pi - A)$ .
Remember the sine subtraction formula: $\sin(x - y) = \sin(x)\cos(y) - \cos(x)\sin(y)$ .

STEP 5

Let's apply this to our expression: $\sin(\pi - A) = \sin(\pi)\cos(A) - \cos(\pi)\sin(A)$

STEP 6

We know that $\sin(\pi) = \textbf{0}$ and $\cos(\pi) = \textbf{-1}$ .
Substituting these values, we get: $\sin(\pi - A) = (\textbf{0})\cos(A) - (\textbf{-1})\sin(A) = \sin(A)$ So, option B is equal to $\sin(A)$ .

STEP 7

We're looking at $\cos(90^\circ - A)$ .
This reminds us of the cofunction identity: $\cos(90^\circ - A) = \sin(A)$ .
So, option C is equal to $\sin(A)$ .

STEP 8

We have $\frac{1}{\csc(A)}$ .
Remember that $\csc(A)$ is the reciprocal of $\sin(A)$ , meaning $\csc(A) = \frac{1}{\sin(A)}$ .

STEP 9

Therefore, $\frac{1}{\csc(A)} = \frac{1}{\frac{1}{\sin(A)}} = \sin(A)$ .
So, option D is equal to $\sin(A)$ .

STEP 10

Options B, C, and D are all equal to $\sin(A)$ .
Only option A, which simplifies to $-\sin(A)$ , is not equal to $\sin(A)$ .
Thus, the answer is A.

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QuestionEach of the following are equal to Sin⁡(A)\operatorname{Sin}(A)Sin(A) EXCEPTA sin⁡(−A)\sin (-A)sin(−A) B sin⁡(π−A)\sin (\pi-A)sin(π−A) (C) cos⁡(90∘−A)\cos \left(90^{\circ}-A\right)cos(90∘−A)D 1csc⁡A\frac{1}{\csc A}cscA1​SUBMIT ANSWER

Studdy Solution

STEP 1

What is this asking? Which of these expressions is *not* equal to sin⁡(A)\sin(A)sin(A)? Watch out! Remember your trigonometric identities and the relationships between sine and cosine!Also, be careful with signs and angles.

STEP 2

1. Analyze option A2. Analyze option B3. Analyze option C4. Analyze option D

STEP 3

We're given sin⁡(−A)\sin(-A)sin(−A).The sine function has a special property: it's an **odd function**!This means sin⁡(−A)=−sin⁡(A)\sin(-A) = -\sin(A)sin(−A)=−sin(A).So, option A is *not* equal to sin⁡(A)\sin(A)sin(A).

STEP 4

We have sin⁡(π−A)\sin(\pi - A)sin(π−A).Remember the **sine subtraction formula**: sin⁡(x−y)=sin⁡(x)cos⁡(y)−cos⁡(x)sin⁡(y)\sin(x - y) = \sin(x)\cos(y) - \cos(x)\sin(y)sin(x−y)=sin(x)cos(y)−cos(x)sin(y).

STEP 5

Let's apply this to our expression: sin⁡(π−A)=sin⁡(π)cos⁡(A)−cos⁡(π)sin⁡(A) \sin(\pi - A) = \sin(\pi)\cos(A) - \cos(\pi)\sin(A) sin(π−A)=sin(π)cos(A)−cos(π)sin(A)

STEP 6

STEP 7

We're looking at cos⁡(90∘−A)\cos(90^\circ - A)cos(90∘−A).This reminds us of the **cofunction identity**: cos⁡(90∘−A)=sin⁡(A)\cos(90^\circ - A) = \sin(A)cos(90∘−A)=sin(A).So, option C *is* equal to sin⁡(A)\sin(A)sin(A).

STEP 8

We have 1csc⁡(A)\frac{1}{\csc(A)}csc(A)1​.Remember that csc⁡(A)\csc(A)csc(A) is the **reciprocal** of sin⁡(A)\sin(A)sin(A), meaning csc⁡(A)=1sin⁡(A)\csc(A) = \frac{1}{\sin(A)}csc(A)=sin(A)1​.

STEP 9

Therefore, 1csc⁡(A)=11sin⁡(A)=sin⁡(A)\frac{1}{\csc(A)} = \frac{1}{\frac{1}{\sin(A)}} = \sin(A)csc(A)1​=sin(A)1​1​=sin(A).So, option D *is* equal to sin⁡(A)\sin(A)sin(A).

STEP 10

Options B, C, and D are all equal to sin⁡(A)\sin(A)sin(A).Only option A, which simplifies to −sin⁡(A)-\sin(A)−sin(A), is *not* equal to sin⁡(A)\sin(A)sin(A).Thus, the answer is A.

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QuestionEach of the following are equal to $\operatorname{Sin}(A)$ EXCEPT
A $\sin (-A)$ B $\sin (\pi-A)$ (C) $\cos \left(90^{\circ}-A\right)$
D $\frac{1}{\csc A}$
SUBMIT ANSWER

What is this asking? Which of these expressions is not equal to $\sin(A)$ ? Watch out! Remember your trigonometric identities and the relationships between sine and cosine!
Also, be careful with signs and angles.

1. Analyze option A
2. Analyze option B
3. Analyze option C
4. Analyze option D

We're given $\sin(-A)$ .
The sine function has a special property: it's an odd function!
This means $\sin(-A) = -\sin(A)$ .
So, option A is not equal to $\sin(A)$ .

We have $\sin(\pi - A)$ .
Remember the sine subtraction formula: $\sin(x - y) = \sin(x)\cos(y) - \cos(x)\sin(y)$ .

Let's apply this to our expression: $\sin(\pi - A) = \sin(\pi)\cos(A) - \cos(\pi)\sin(A)$

We're looking at $\cos(90^\circ - A)$ .
This reminds us of the cofunction identity: $\cos(90^\circ - A) = \sin(A)$ .
So, option C is equal to $\sin(A)$ .

We have $\frac{1}{\csc(A)}$ .
Remember that $\csc(A)$ is the reciprocal of $\sin(A)$ , meaning $\csc(A) = \frac{1}{\sin(A)}$ .

Therefore, $\frac{1}{\csc(A)} = \frac{1}{\frac{1}{\sin(A)}} = \sin(A)$ .
So, option D is equal to $\sin(A)$ .

Options B, C, and D are all equal to $\sin(A)$ .
Only option A, which simplifies to $-\sin(A)$ , is not equal to $\sin(A)$ .
Thus, the answer is A.