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kaheart [24]
3 years ago
7

Use the identities for the cosine of a sum or a difference to write the expression as a single function of x.

Mathematics
1 answer:
lions [1.4K]3 years ago
3 0

Answer:

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Step-by-step explanation:


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Help me I need to get a 100
swat32

Answer:

option 2

Step-by-step explanation:

3 0
3 years ago
PLEASE HELP ITS A TEST<br> Solve for x : - 1/2 (x + 3) - 10 = -6.5
Ksju [112]

Answer hey:

Step-by-step explanation:

4 0
3 years ago
Please prove this........​
Crazy boy [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π    →     C = π - (A + B)

                                    → sin C = sin(π - (A + B))       cos C = sin(π - (A + B))

                                    → sin C = sin (A + B)              cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS:                        (sin 2A + sin 2B) + sin 2C

\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C

\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C

\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C

\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)

\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]

\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B

\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C    \checkmark

7 0
3 years ago
Solve the initial value problem: y'(x)=(4y(x)+25)^(1/2) ,y(1)=6. you can't really tell, but the '1/2' is the exponent
goblinko [34]

Answer:

y(x)=x^2+5x

Step-by-step explanation:

Given: y'=\sqrt{4y+25}

Initial value: y(1)=6

Let y'=\dfrac{dy}{dx}

\dfrac{dy}{dx}=\sqrt{4y+25}

Variable separable

\dfrac{dy}{\sqrt{4y+25}}=dx

Integrate both sides

\int \dfrac{dy}{\sqrt{4y+25}}=\int dx

\sqrt{4y+25}=2x+C

Initial condition, y(1)=6

\sqrt{4\cdot 6+25}=2\cdot 1+C

C=5

Put C into equation

Solution:

\sqrt{4y+25}=2x+5

or

4y+25=(2x+5)^2

y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}

y(x)=x^2+5x

Hence, The solution is y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4} or y(x)=x^2+5x

4 0
3 years ago
Find a formula for the nth term of the arithmetic sequence a1=-17 a2=-12 a3=-7 a4=-2
MAVERICK [17]

Answer:

Step-by-step explanation:

Recursive formula

a1 = -17

a^n = a^n-1 + 5

Explicit formula:

a^n = -17 + (n-1) * 5

3 0
3 years ago
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