Question

need urgent help asap! trig question 7, 9, 10, 11, 12plz,, plz show step by step work on each one plz!​

Answer 1

You just need a calculator for these so there is no working

7)
`4.569245397*10^(-3)`

8)
`(1)/(2)`

9)0.5773502692

10)
`6.853838284*10^(-3)`

I am not sure for the rest, sorry havent done that far in school yet :)

Hope these helps :)

Answer 2

You need the Unit Circle to answer these questions! (attached)

7. Answer:
`\bold{(\sqrt6-\sqrt2)/(4)}`

Explanation:

`sin(\pi)/(12)=15^o\\\\\text{Since }15^o\text{ is not on the Unit Circle, you will need to use the difference}\\\text{formula for sin: sin(A - B) = sinAcosB - cosAsinB}\\\\sin15=sin(45-30)=(\sqrt2)/(2)\cdot(\sqrt3)/(2)-(\sqrt2)/(2)\cdot(1)/(2)=(\sqrt6)/(4)-(\sqrt2)/(4)=\boxed{(\sqrt6-\sqrt2)/(4)}`

8. Answer: 0

Explanation:

The sum formula for cos is: cos (A + B) = cosA cosB - sinA sinB

Since we are given cos75° cos15° - sin75° sin15°, we can conclude that we are looking for cos (75° + 15°) = cos 90°.

The Unit Circle shows us that cos 90° = 0

9. Answer:
`\bold{(\sqrt3)/(3)}`

Explanation:

The given expression is for tan (A + B).

tan (10° + 20°) = tan 30° =
`(sin30^o)/(cos30^o)=(1)/(\sqrt3)\implies\boxed{(\sqrt3)/(3)}`

10. Answer:
`\bold{\pm(2-\sqrt2)/(2)}`

Explanation:

`sin(\pi)/(8)=22.5^o=sin(45^o)/(2)\\\text{So we need to use the half-angle formula for sin:}\\\\sin(\theta)/(2)=\pm\sqrt(1-cos\theta)/(2)}\\\\sin(45^o)/(2)=\pm\sqrt(1-cos45^o)/(2)}=\pm\sqrt(1-(\sqrt2)/(2))/(2)}=\pm\sqrt((2)/(2)-(\sqrt2)/(2))/((4)/(2))}=\boxed{\pm(2-\sqrt2)/(2)}`

11. TOO BLURRY TO SEE THE PROBLEM

12. Answer:
`\bold{(7)/(25)}`

Explanation:

Quadrant II means that sin is positive and cos is negative.

`sin\ x=(3)/(5)\implies cos\ x=-(4)/(5)\\\\cos\ 2x=cos^2\ x-sin^2\ x\\\\.\qquad=\bigg((-4)/(5)\bigg)^2-\bigg((3)/(5)\bigg)^2\\\\.\qquad=(16)/(25)-(9)/(25)\\\\.\qquad=\boxed{(7)/(25)}`