Answer:
sin(θ+β)=−
5
7
−4
15
2
step by step explanation;
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