+0

# math help

0
139
6

1. Right triangle has legs with lengths $$\sqrt7$$ on the bottom and 1 on the right side. Use this information to find:

\begin{align*} &\sqrt{8}\cdot \cos\left(\arctan\left(\sqrt{7}\right)\right), \\ &\sqrt{8} \cdot \cos\left(\arccos\left(\frac{1}{\sqrt{8}}\right)\right), \\ &\sqrt{8} \cdot \sin\left(\arccos\left(\frac{1}{\sqrt{8}}\right)\right), \end{align*}in the same order.

2. Calculate $$\arcsin(1), \arccos\left(\frac{1}{2} \right), \arctan(1), \arcsin\left(-\frac{1}{2}\right)$$ in radians and enter them below.

3. A 3-4-5 right triangle has length 3 on the bottom side and 4 on the right side. Use this information to find:

$$\sin\left(\arctan\left(\frac{3}{4}\right)\right), \cos\left(\arcsin\left(\frac{3}{5}\right)\right), \cot\left(\arctan\left(\frac{4}{3}\right)\right), \sin\left(\arcsin\left(\frac{4}{5}\right)\right)$$in the same order.

Thank you for your help! I'm having a bit of trouble with calculus.

Sep 12, 2019

#1
+3

1. √8  * cos [ arctan (√7 ]

This says that   we first need to find the cosine  of an angle whose tangent  = y / x  =  √7/1

So we have   x / √[y^2 + x^2 ] =      1  /  √[ (√7)^2 + 1^2]    =   1 /  √ [7 + 1]   =   1  / √8

So...

√8  *  1 / √8   =      1

√8  *  cos  [ arccos (1/ √8) ]

Note that  cos  [ arccos ( 1/√8) ]    =    1/√8

So, again.... √8 * 1 / √8    =    1

√8 *  sin [ arcos (1/√8)].....this says that we first need to find the sin of an angle whose cosine  = 1 / √8

y  =   √[ r^2 - x^2]   =  √[ 8 - 1 ] = √7

So  sin [ arccos (1 / √8 ) ] =  y/r  =  √7/√8

So

√8  *  √7/√8    =    √7   Sep 12, 2019
#5
0

Thank you so much for your work! I really appreciate it :)

Guest Sep 13, 2019
#2
+3

2.

arcsin 1   =   we are asking.....where is the sine = 1 ???......this is  at  pi/2 rads

arccos 1/2    =   where is the cosine = 1/2 ???.....this is at   pi/3 rads

arctan 1  =  where is the tangent  = 1  ???......this is at    pi/4 rads

arcsin -1/2   =  where is the sine  = -1/2 ???.....this is at  - pi/6  rads

P.S. - you should memorize the values of the sine, cosine and tangent  at  0, pi/6, pi/4, pi/3  and pi/2 rads   Sep 12, 2019
#3
0

Here is a sample of how to solve #3 see the image....think you can now do the others? Sep 12, 2019
#6
0

Thank you for your help as well, especially for the hint!

Guest Sep 13, 2019
#4
+3

3.

Note that  r =  √[3^2 + 4^2]  =  √25   =   5

sin [ arctan (3/4) ]   =  we are looking for the sine of an angle whose tangent = 3/4  =     y  r  =     3 / 5

cos [arcsin (3/5)] =  we are looking for the cosine whose sine  = 3/5   =  [ 5^2 - 3^2] / 5  =  x/r =  √16/5   = 4/5

cot [ arctan ( 4/3) ] =   we are looking for the tangent of an angle whose cot  = 4/3  =   3/4

sin [arcsin (4/5) ]    =  4/5   Sep 12, 2019