Question

Please help me with this question

Answer 1

`\large\boxed{Q2.\ \cos\theta=(3)/(5)}\\\boxed{Q3.\ \sin(\alpha+\beta)=-(13)/(85)}`

Explanation:

Q2.

`270^o<\theta<360^o-IV\ Quarter\\\\\sin\theta=-(4)/(5),\ \cos\theta=?\\\\\text{Use}\ \sin^2x+\cos^2x=1\\\\\left(-(4)/(5)\right)^2+\cos^2\theta=1\\\\(16)/(25)+\cos^2\theta=1\qquad\text{subtract}\ (16)/(25)\ \text{from both sides}\\\\\cos^2\theta=(25)/(25)-(16)/(25)\\\\\cos^2\theta=(9)/(25)\to\cos\theta=\pm\sqrt{(9)/(25)}\\\\\cos\theta=\pm(3)/(5)\\\\\theta\ \text{is in IV Quarter. Therefore}\ \cos\theta>0\to\boxed{\cos\theta=(3)/(5)}`

Q3.

`180^o\alpha<270^o-III\ Quarter\\\\\cos\alpha=-(8)/(17),\\\\\sin^2\alpha+\left(-(8)/(17)\right)^2=1\\\\\sin^2\alpha+(64)/(289)=1\qquad\text{subtract}\ (64)/(289)\ \text{from both sides}\\\\\sin^2\alpha=(289)/(289)-(64)/(289)\\\\\sin^2\alpha=(225)/(289)\to\sin\alpha=\pm\sqrt{(225)/(289)}\\\\\sin\alpha=\pm(15)/(17)\\\\\alpha\ \text{is in III Quarter. Therefore}\ \sin\alpha=-(15)/(17)`

`270^o<\beta<360^o-IV\ Quarter\\\\\sin\beta=-(4)/(5)\\\\\left(-(4)/(5)\right)^2+\cos^2\beta=1\\\\(16)/(25)+\cos^2\beta=1\qquad\text{subtract}\ (16)/(25)\ \text{from both sides}\\\\\cos^2\beta=(25)/(25)-(16)/(25)\\\\\cos^2\beta=(9)/(25)\to\cos\beta=\pm\sqrt{(9)/(25)}\\\\\cos\beta=\pm(3)/(5)\\\\\beta\ \text{is in IV Quarter. Therefore}\ \cos\beta=(3)/(5)`

`\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha\\\\\text{Substitute:}\\\\\sin(\alpha+\beta)=\left(-(15)/(17)\right)\left((3)/(5)\right)+\left(-(4)/(5)\right)\left(-(8)/(17)\right)\\\\=-(45)/(85)+(32)/(85)=-(13)/(85)`