Question

Help please it's number lines​

Answer 1

sin(θ+β)=−

5

7

−4

15

2

Step-by-step explanation:

step 1

Find the sin(\theta)sin(θ)

we know that

Applying the trigonometric identity

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

2

(θ)+cos

2

(θ)=1

we have

cos(\theta)=-\frac{\sqrt{2}}{3}cos(θ)=−

3

2

substitute

sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1sin

2

(θ)+(−

3

2

)

2

=1

sin^2(\theta)+ \frac{2}{9}=1sin

2

(θ)+

9

2

=1

sin^2(\theta)=1- \frac{2}{9}sin

2

(θ)=1−

9

2

sin^2(\theta)= \frac{7}{9}sin

2

(θ)=

9

7

sin(\theta)=\pm\frac{\sqrt{7}}{3}sin(θ)=±

3

7

Remember that

π≤θ≤3π/2

so

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

sin(\theta)=-\frac{\sqrt{7}}{3}sin(θ)=−

3

7

step 2

Find the sec(β)

Applying the trigonometric identity

tan^2(\beta)+1= sec^2(\beta)tan

2

(β)+1=sec

2

(β)

we have

tan(\beta)=\frac{4}{3}tan(β)=

3

4

substitute

(\frac{4}{3})^2+1= sec^2(\beta)(

3

4

)

2

+1=sec

2

(β)

\frac{16}{9}+1= sec^2(\beta)

9

16

+1=sec

2

(β)

sec^2(\beta)=\frac{25}{9}sec

2

(β)=

9

25

sec(\beta)=\pm\frac{5}{3}sec(β)=±

3

5

we know

0≤β≤π/2 ----> II Quadrant

so

sec(β), sin(β) and cos(β) are positive

sec(\beta)=\frac{5}{3}sec(β)=

3

5

Remember that

sec(\beta)=\frac{1}{cos(\beta)}sec(β)=

cos(β)

1

therefore

cos(\beta)=\frac{3}{5}cos(β)=

5

3

step 3

Find the sin(β)

we know that

tan(\beta)=\frac{sin(\beta)}{cos(\beta)}tan(β)=

cos(β)

sin(β)

we have

tan(\beta)=\frac{4}{3}tan(β)=

3

4

cos(\beta)=\frac{3}{5}cos(β)=

5

3

substitute

(4/3)=\frac{sin(\beta)}{(3/5)}(4/3)=

(3/5)

sin(β)

therefore

sin(\beta)=\frac{4}{5}sin(β)=

5

4

step 4

Find sin(θ+β)

we know that

sin(A + B) = sin A cos B + cos A sin Bsin(A+B)=sinAcosB+cosAsinB

so

In this problem

sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)sin(θ+β)=sin(θ)cos(β)+cos(θ)sin(β)

we have

sin(\theta)=-\frac{\sqrt{7}}{3}sin(θ)=−

3

7

cos(\theta)=-\frac{\sqrt{2}}{3}cos(θ)=−

3

2

sin(\beta)=\frac{4}{5}sin(β)=

5

4

cos(\beta)=\frac{3}{5}cos(β)=

5

3

substitute the given values in the formula

sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})sin(θ+β)=(−

3

7

)(

5

3

)+(−

3

2

)(

5

4

)

sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})sin(θ+β)=(−3

15

7

)+(−4

15

2

)

sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}sin(θ+β)=−

5

7

−4

15

2

Step-by-step explanation:

i hope it helps to you