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lesya692 [45]
3 years ago
6

Smart Start Review :

Mathematics
1 answer:
Delvig [45]3 years ago
6 0

Answer:

A. 5W-21 B. -60+6W C. -7+2W

Step-by-step explanation:

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Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))
Sonja [21]

Answer:

\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}

Let a=\sin^{-1}(\frac{2}{3}).

With some restriction on a this means:

\sin(a)=\frac{2}{3}

We need to find \tan(a).

\sin^2(a)+\cos^2(a)=1 is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

(\frac{2}{3})^2+\cos^2(a)=1

\frac{4}{9}+\cos^2(a)=1

Subtract 4/9 on both sides:

\cos^2(a)=\frac{5}{9}

Take the square root of both sides:

\cos(a)=\pm \sqrt{\frac{5}{9}}

\cos(a)=\pm \frac{\sqrt{5}}{3}

The cosine value is positive because a is a number between -\frac{\pi}{2} and \frac{\pi}{2} because that is the restriction on sine inverse.

So we have \cos(a)=\frac{\sqrt{5}}{3}.

This means that \tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}.

Multiplying numerator and denominator by 3 gives us:

\tan(a)=\frac{2}{\sqrt{5}}

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

\tan(a)=\frac{2\sqrt{5}}{5}

Let's continue on to letting b=\cos^{-1}(\frac{1}{7}).

Let's go ahead and say what the restrictions on b are.

b is a number in between 0 and \pi.

So anyways b=\cos^{-1}(\frac{1}{7}) implies \cos(b)=\frac{1}{7}.

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of b.

\cos^2(b)+\sin^2(b)=1

(\frac{1}{7})^2+\sin^2(b)=1

\frac{1}{49}+\sin^2(b)=1

Subtract 1/49 on both sides:

\sin^2(b)=\frac{48}{49}

Take the square root of both sides:

\sin(b)=\pm \sqrt{\frac{48}{49}

\sin(b)=\pm \frac{\sqrt{48}}{7}

\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}

\sin(b)=\pm \frac{4\sqrt{3}}{7}

So since b is a number between 0 and \pi, then sine of this value is positive.

This implies:

\sin(b)=\frac{4\sqrt{3}}{7}

So \tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}.

Multiplying both top and bottom by 7 gives:

\frac{4\sqrt{3}}{1}= 4\sqrt{3}.

Let's put everything back into the first mentioned identity.

\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}

\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}

\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}

4 0
3 years ago
Uhhhh anybody know this lsssss help ims tufff
Sever21 [200]
For 9. the third table and for 10. the second table
4 0
3 years ago
Round to the nearest tenth, I will give brainlyest PLEASE
BabaBlast [244]

1. cos B= a/x

x= 11/cos 37°

x= 13.77

2. tan A= a/x

x= 4/tan 41°

x= 4.60

3. sin B = x/c

x= sin 37° × 10.3

x= 6.20

4. tan A= a/b

tan A= 4/13

A= tan-1(0.31)

A= 17°6'10"

5. cos A= b/c

cos A= 12/13

A= cos -¹(0.92)

A= 22°37'12"

6. tan B= a/b

tan B= 3/3

tan B= 1

B= tan‐¹(1)

B= 45°

7 0
2 years ago
If we recall that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side, which
Inga [223]

Answer:

Step-by-step explanation:

the first ine make sense to me

if not right sorry

7 0
3 years ago
5 + x / 4 = 1 what is x?
Ratling [72]
X/4= -4
x= -16.......
4 0
3 years ago
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