Math / Trigonometry

QuestionSimplify the expression below in terms of $p$ and $q$ . Your final answer should contain no trigonometric functions and should be written as an expression in $p$ and $q$ solely. $\sin (\arcsin (p)+\arccos (q))$

Studdy Solution

STEP 1

1. The expression involves inverse trigonometric functions.
2. The range of $\arcsin$ is $[- \frac{\pi}{2}, \frac{\pi}{2}]$ and the range of $\arccos$ is $[0, \pi]$ .
3. We are given that $\sin(\theta) = p$ and $\cos(\phi) = q$ for some angles $\theta$ and $\phi$ .
4. We need to simplify the expression using trigonometric identities.

STEP 2

1. Express $\arcsin(p)$ and $\arccos(q)$ in terms of angles.
2. Use the angle addition formula for sine.
3. Simplify the resulting expression in terms of $p$ and $q$ .

STEP 3

Let $\theta = \arcsin(p)$ and $\phi = \arccos(q)$ . Then, $\sin(\theta) = p$ and $\cos(\phi) = q$ .

STEP 4

Use the angle addition formula for sine:
$\sin(\theta + \phi) = \sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi)$

STEP 5

Substitute the known values:
$\sin(\theta + \phi) = p \cdot q + \cos(\theta) \cdot \sin(\phi)$

STEP 6

Find $\cos(\theta)$ using the identity $\cos^2(\theta) + \sin^2(\theta) = 1$ :
$\cos(\theta) = \sqrt{1 - p^2}$

STEP 7

Find $\sin(\phi)$ using the identity $\sin^2(\phi) + \cos^2(\phi) = 1$ :
$\sin(\phi) = \sqrt{1 - q^2}$

STEP 8

Substitute $\cos(\theta)$ and $\sin(\phi)$ back into the expression:
$\sin(\theta + \phi) = p \cdot q + \sqrt{1 - p^2} \cdot \sqrt{1 - q^2}$
The simplified expression in terms of $p$ and $q$ is:
$\boxed{p \cdot q + \sqrt{1 - p^2} \cdot \sqrt{1 - q^2}}$

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QuestionSimplify the expression below in terms of $p$ and $q$ . Your final answer should contain no trigonometric functions and should be written as an expression in $p$ and $q$ solely. $\sin (\arcsin (p)+\arccos (q))$

Studdy Solution

STEP 1

STEP 2

1. Express $\arcsin(p)$ and $\arccos(q)$ in terms of angles.
2. Use the angle addition formula for sine.
3. Simplify the resulting expression in terms of $p$ and $q$ .

STEP 3

Let $\theta = \arcsin(p)$ and $\phi = \arccos(q)$ . Then, $\sin(\theta) = p$ and $\cos(\phi) = q$ .

STEP 4

Use the angle addition formula for sine:
$\sin(\theta + \phi) = \sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi)$

STEP 5

Substitute the known values:
$\sin(\theta + \phi) = p \cdot q + \cos(\theta) \cdot \sin(\phi)$

STEP 6

Find $\cos(\theta)$ using the identity $\cos^2(\theta) + \sin^2(\theta) = 1$ :
$\cos(\theta) = \sqrt{1 - p^2}$

STEP 7

Find $\sin(\phi)$ using the identity $\sin^2(\phi) + \cos^2(\phi) = 1$ :
$\sin(\phi) = \sqrt{1 - q^2}$

STEP 8

Substitute $\cos(\theta)$ and $\sin(\phi)$ back into the expression:
$\sin(\theta + \phi) = p \cdot q + \sqrt{1 - p^2} \cdot \sqrt{1 - q^2}$
The simplified expression in terms of $p$ and $q$ is:
$\boxed{p \cdot q + \sqrt{1 - p^2} \cdot \sqrt{1 - q^2}}$

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QuestionSimplify the expression below in terms of ppp and qqq. Your final answer should contain no trigonometric functions and should be written as an expression in ppp and qqq solely. sin⁡(arcsin⁡(p)+arccos⁡(q))\sin (\arcsin (p)+\arccos (q))sin(arcsin(p)+arccos(q))

Studdy Solution

STEP 1

STEP 2

1. Express arcsin⁡(p)\arcsin(p)arcsin(p) and arccos⁡(q)\arccos(q)arccos(q) in terms of angles.2. Use the angle addition formula for sine.3. Simplify the resulting expression in terms of ppp and qqq.

STEP 3

Let θ=arcsin⁡(p)\theta = \arcsin(p)θ=arcsin(p) and ϕ=arccos⁡(q)\phi = \arccos(q)ϕ=arccos(q). Then, sin⁡(θ)=p\sin(\theta) = psin(θ)=p and cos⁡(ϕ)=q\cos(\phi) = qcos(ϕ)=q.

STEP 4

Use the angle addition formula for sine:sin⁡(θ+ϕ)=sin⁡(θ)cos⁡(ϕ)+cos⁡(θ)sin⁡(ϕ)\sin(\theta + \phi) = \sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi)sin(θ+ϕ)=sin(θ)cos(ϕ)+cos(θ)sin(ϕ)

STEP 5

Substitute the known values:sin⁡(θ+ϕ)=p⋅q+cos⁡(θ)⋅sin⁡(ϕ)\sin(\theta + \phi) = p \cdot q + \cos(\theta) \cdot \sin(\phi)sin(θ+ϕ)=p⋅q+cos(θ)⋅sin(ϕ)

STEP 6

Find cos⁡(θ)\cos(\theta)cos(θ) using the identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1cos2(θ)+sin2(θ)=1:cos⁡(θ)=1−p2\cos(\theta) = \sqrt{1 - p^2}cos(θ)=1−p2​

STEP 7

Find sin⁡(ϕ)\sin(\phi)sin(ϕ) using the identity sin⁡2(ϕ)+cos⁡2(ϕ)=1\sin^2(\phi) + \cos^2(\phi) = 1sin2(ϕ)+cos2(ϕ)=1:sin⁡(ϕ)=1−q2\sin(\phi) = \sqrt{1 - q^2}sin(ϕ)=1−q2​

STEP 8

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QuestionSimplify the expression below in terms of $p$ and $q$ . Your final answer should contain no trigonometric functions and should be written as an expression in $p$ and $q$ solely. $\sin (\arcsin (p)+\arccos (q))$

1. Express $\arcsin(p)$ and $\arccos(q)$ in terms of angles.
2. Use the angle addition formula for sine.
3. Simplify the resulting expression in terms of $p$ and $q$ .

Let $\theta = \arcsin(p)$ and $\phi = \arccos(q)$ . Then, $\sin(\theta) = p$ and $\cos(\phi) = q$ .

Use the angle addition formula for sine:
$\sin(\theta + \phi) = \sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi)$

Substitute the known values:
$\sin(\theta + \phi) = p \cdot q + \cos(\theta) \cdot \sin(\phi)$

Find $\cos(\theta)$ using the identity $\cos^2(\theta) + \sin^2(\theta) = 1$ :
$\cos(\theta) = \sqrt{1 - p^2}$

Find $\sin(\phi)$ using the identity $\sin^2(\phi) + \cos^2(\phi) = 1$ :
$\sin(\phi) = \sqrt{1 - q^2}$