Question

give the exact value of tan(arccos √3/2 + arcsin -3/5.

Answer 1

what is that is it like this \(\large \rm tan(cos^{-1}(\sqrt{3}/2)+sin^{-1}(-3/5))\)
everything inside tan? or not

Answer 2

yes how you have it written

Answer 3

Well, then I would use the sum of tangents formula to start off with
\[\tan(a + b) = \frac{ tana + \tan b }{ 1- tana \cdot tanb }\]

Alternatively, you could use the division of the sum formulas for sin and cos (since values would cancel out)

\[\frac{ \sin(a + b) }{ \cos(a+ b) } = \frac{ (sina)(cosb) + (cosa) (sinb) }{ (cosa) (cosb) - (sina)( sinb) }\]

Answer 4

you need to ask you self what angle gives \(\sqrt{3}/2\) whe you cosine it
also the same for -3/5 what angle gives that ratio

Answer 5

so arccos=tan a and arcsin=tanb???

Answer 6

arccos = a, arcsin = b. So you would have:
\[\frac{ \tan(\cos^{-1}(\sqrt{3}/2)) + \tan(\sin^{-1}(-3/5)) }{ 1-\tan(\cos^{-1}(\sqrt{3}/2))\cdot \tan(\sin^{-1}(-3/5)) }\]

For the cosine inverse portions, you can use your knowledge of what angle gives \(\sqrt{3}/2\) when you take the cosine of it (or check a unit circle), or we could use triangles for all of these values. We need a triangle for the inverse sine portions anyway :P

Answer 7

yes you need some knowledge about some angles and their sin cos
and rational trigonometry

Answer 8

Having \(tan(sin^{-1}(-3/5))\) can be translated as something along the lines of
"Some triangle has a sine valueof -3/5. What is the tangent of this same triangle?" Using tht idea, you can solve the \(tan(sin^{-1}(-3/5)\) portion of your fraction. Unit circle and knowledge of it works out just fine for the other values.

Answer 9

4

Answer 10

Right. So if that missing value is 4, what's the tangent of that triangle?

Answer 11

using Pythagoras you solve for the missing side and then find the tangent

Answer 12

correct it is 4

Answer 13

-3/4

Answer 14

Tan is OPPOSITE/ADJACENT!

Answer 15

yep

Answer 16

Bingo. So -3/4 can now replace all of the \(tan(sin^{-1}(-3/5))\) portions of our formula. Now we just need the values for \(tan(cos^{-1}(\sqrt{3}/2))\)

Answer 17

square root 3

Answer 18

look here
http://www.ambrsoft.com/equations/Trigonometric_files/sinvalues.PNG
but you need to know this values, you can't do trig without them

Answer 19

no square root 3 over 3

Answer 20

no you need to find the angle whose sin is root3/2

Answer 21

that is 60 degrees

Answer 22

\[\tan(\cos^{-1}(\sqrt{3}/2)) = \tan(\frac{ \pi }{ 6 }) = \frac{ 1/2 }{ \sqrt{3}/2 } = \frac{ 1 }{ \sqrt{3} }= \frac{ \sqrt{3} }{ 3 }\]

Answer 23

i mean cosine sorry not sin

Answer 24

so now i am going to replace all of the cos values with square root 3 over 3?

Answer 25

We now have what we need, so let's plug in the values we found :P
\[\frac{ \frac{ \sqrt{3} }{ 3 }- \frac{ 3 }{ 4 } }{ 1- \frac{ \sqrt{3} }{ 3 } \cdot \frac{ -3 }{ 4 }}\]

You'll want to simplify this now.

Answer 26

Able to simplify that ok? :)

Answer 27

workin on it :) i find the LCD for the top?

Answer 28

That works. You can actually find the common denominator for all the terms. If the denominators of every term in the entire problem are 1, 3, and 4, what would that common denominator be?

Answer 29

12

Answer 30

Yep. So you can actually multiply everything, top and bottom, by 12. That would give you:
\[\frac{ 4\sqrt{3}-9 }{ 12 + 3\sqrt{3} }\]
I got rid of the minus sign on bottom, there were two negative signs that could cancel. BUt I multiplied every term by 12 as a common denominator and simplified the fraction substantially.

Now, normally, they would want you to rationalize the denominator to remove any square roots. WOuld you kno how to do that?

Answer 31

ya divide the top and bottom by square root 3

Answer 32

Well, you need to multiply top and bottom by the conjugate actually. Do you know what the conjugate of the denominator would be?

Answer 33

12-√3/3?

Answer 34

I unfortunately have to head out. But once you multiply top and bottom by \(12-3\sqrt{3}\) and simplify, you'll have your final answer. Good luck ^_^

Answer 35

thanks for ur help!

Answer 36

NP :3

Answer 37

\[\tan \left(\sin ^{-1}\left(-\frac{3}{5}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)=\frac{1}{39} \left(25 \sqrt{3}-48\right)=-0.12048 \]