Question

6. Find an exact value. (1 point)
sin 75°


a) quantity negative square root of six plus square root of two divided by four.
b) quantity square root of six minus square root of two divided by four.
c) quantity negative square root of six minus square root of two divided by four.
d) quantity square root of six plus square root of two divided by four.

7. Find an exact value. (1 point)
sine of negative eleven pi divided by twelve.


a) quantity square root of six plus square root of two divided by four.
b) quantity negative square root of six minus square root of two divided by four.
c) quantity square root of two minus square root of six divided by four.
d) quantity square root of six minus square root of two divided by four.

8. Write the expression as the sine, cosine, or tangent of an angle. (1 point)
sin 9x cos x - cos 9x sin x


a) sin 10x
b) cos 8x
c) sin 8x
d) cos 10x

9. Write the expression as the sine, cosine, or tangent of an angle. (1 point)
cos 112° cos 45° + sin 112° sin 45°


a) sin 157°
b) sin 67°
c) cos 157°
d) cos 67°

10. Rewrite with only sin x and cos x. (1 point)
sin 2x - cos 2x


a) 2 sinx cosx - 1 + 2 sin2x
b) 2 sin x cos2x - 1 + 2 sin2x
c) 2 sin x cos2x - sin x + 1 - 2 sin2x
d) 2 sin x cos2x - 1 - 2 sin2x


PLEASE HELP!

Answer 1

6 Find an exact value. 
sin 75°
sin(A+B)=sin(A)cos(B)+cos(A)sin(B)
sin(45)=cos(45)=(2^0.5)/2    sin(30)=0.5      cos(30)=(3^0.5)/2
sin(45+30)=sin(45)cos(30)+cos(45)sin(30)=(6^0.5+2^0.5)/4
the answer is the letter d) quantity square root of six plus square root of two divided by four.

7. Find an exact value. 
sine of negative eleven pi divided by twelve.

sin(-11pi/12) = -sin(11pi/12) = -sin(pi - pi/12) = -sin(pi/12) = -sin( (pi/6) / 2)

= - sqrt( (1-cos(pi/6) ) / 2) = -sqrt( (1-√3/2) / 2 ) = -(√3-1) / 2√2=(√2-√6)/4
the answer is the letter c) quantity square root of two minus square root of six divided by four.

8. Write the expression as the sine, cosine, or tangent of an angle. 
sin 9x cos x - cos 9x sin x

sin(A−B)=sinAcosB−cosAsinB

sin(9x−x)= sin9xcosx−cos9xsinx= sin(8x)

the answer is the letter c) sin 8x

9. Write the expression as the sine, cosine, or tangent of an angle. 
cos 112° cos 45° + sin 112° sin 45°

cos(A−B)=cosAcosB+sinAsinB

cos(112−45)=cos112cos45+sin112sin45=cos(67)

the answer is the letter d) cos 67°

10. Rewrite with only sin x and cos x.

sin 2x - cos 2x

 

sin2x = 2sinxcosx
cos2x = (cosx)^2 - (sinx)^2 = 2(cosx)^2 -1 = 1- 2(sinx)^2

sin2x- cos2x=2sinxcosx-(1- 2(sinx)^2=2sinxcosx-1+2(sinx)^2

sin2x- cos2x=2sinxcosx-1+2(sinx)^2

the answer is the letter b) 2 sin x cos2x - 1 + 2 sin2x

Answer 2

Answer:

6.d.Quantity square root of six plus square root of two divided by four.

7.:Quantity negative square root 2  minus square root three divided by two.

8.C.Sin 8x

9.d. $Cos 67^{\circ}$

10.b.$2sin x cos x-1 +2 sin^2 x$

Step-by-step explanation:

6.$Sin75^{\circ}$=Sin(45+30)

Sin(A+B)=Sin A Cos B+Sin B Cos A

Using identity

Sin(45+30)= Sin 45 Cos 30+ Cos 45 Sin 30=$\frac{1}{\sqrt2}\cdot \frac{\sqrt3}{2}+\frac{1}{\sqrt2}\cdot\frac{1}{2}$

$Sin 45^{\circ}=Cos 45^{\circ}=\frac{1}{\sqrt2}$

$Sin 30=\frac{1}{2},Cos 30=\frac{\sqrt3}{2}$

$Sin(45+30)=\frac{\sqrt3}{2\sqr2}+\frac{1}[2\sqt2}$

$Sin(45+30)=\frac{\sqrt3\times \sqrt2}{2\sqrt2\times\sqrt2}+\frac{\sqrt2}{2\sqrt\times \sqrt2}$

$Sin(45+30)=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}=\frac{\sqrt6+\sqrt2}{4}$

d.Quantity square root of six plus square root of two divided by four.

7.$Sin(-\frac{11\pi}{12})$

$Sin(-\frac{11\pi}{12})=-Sin\frac{11\pi}{12}$

$Sin (-x)=-Sin x$

$Sin (-\frac{11\pi}{12})=-Sin(\pi-\frac{\pi}{12})=-Sin\frac{\pi}{12}$

$-Sin\frac{\frac{\pi}{6}}{2}=-\sqrt{\frac{1-cos\frac{\pi}{6}}{2}}$

=$-\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=-\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=-\frac{\sqrt{2-\sqrt3}}{2}$

$Sin(-\frac{11\pi}{12})=-\frac{\sqrt{2-\sqrt3}}{2}$

Answer :Quantity negative square root 2  minus square root three divided by two.

8.$sin 9x Cos x-Cos 9x sin x$

$sin (A-B)=Sin A Cos B- Sin B Cos A$

Using this identity

Then we get

$Sin 9x Cos x- Sin x Cos 9x= Sin (9x-x)=Sin 8x$

C.Sin 8x

9$.Cos 112^{\circ}Cos 45^{\circ}+ sin 112^{\circ} sin 45 ^{\circ}$

$Cos (A-B)=Cos A Cos B+ Sin A Sin B$

Using this identity then we get

$Cos (112-45)=Cos 67^{\circ}$

d. $Cos 67^{\circ}$

10. sin 2x -cos 2x

$Sin 2 x=2 sin x cos x$

$Cos 2x =1- 2 sin^2 x$

Using above identities

Therefore, $sin 2x- cos 2x=2sin xcos x-1+2 sin^2x$

b.$2sin x cos x-1 +2 sin^2 x$