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Rasek [7]
3 years ago
7

6. Find an exact value. (1 point)

Mathematics
2 answers:
Zepler [3.9K]3 years ago
5 0

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

vladimir2022 [97]3 years ago
4 0
<span>6 Find an exact value. 
sin 75°
</span>sin(A+B)=sin(A)cos(B)+cos(A)sin<span>(B)
</span>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.

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

</span>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.

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

sin(A−B)=sinAcosB−cosAsinB

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

the answer is the letter c) sin 8x

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

</span>

cos(A−B)=cosAcosB<span>+sinA</span>sinB

cos(112−45)=cos112cos45<span>+sin112</span>sin45=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<span>
cos2x = (cosx)^2 - (sinx)^2 = 2(cosx)^2 -1 = 1- 2(sinx)^2</span>

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

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

<span>the answer is the letter <span>b) 2 sin x cos2x - 1 + 2 sin2x</span></span>
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