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Pre calc help! I did some questions but I don’t understand how to do the others!! :(

Guest Oct 23, 2018
 #1
avatar+90969 
+1

tan  ( arccos (2/3) )

 

We are looking for the tangent  of an angle whose cosine is  2/3

 

So..we know  x / r   and we are looking for y/ x...so we have

 

sqrt  ( 3^2 - 2^2)  =  sqrt 5   =  y

 

So....y / x  =     sqrt(5) / 2

 

 

cool cool cool

CPhill  Oct 24, 2018
 #2
avatar+90969 
+1

cos ( arcsin (-3/5) )

 

We are looking for the cosine .... (x / r)..... of an angle whose sin  is  -3/5  =  y / r

 

So  we need to find x

 

sqrt [ 5^2  - (-3)^2  ] = sqrt [ 25 - 9 ] =  4  = x

 

This angle is in the 4th quadrant....the cosine is positive here

 

So...this evaluates to  x / r  =   4/5

 

 

cool cool cool

CPhill  Oct 24, 2018
 #3
avatar+90969 
+1

sec (arcos (x) )

 

The cosine of this angle =  x / 1

 

So..the sec  is the reciprocal,  1 / x

 

 

 

cot  ( arccos (3x) )

 

The cosine of this angle  =   [3x] / 1    =  3x

 

We are looking for the  cot =    cos / sine

 

So...the sine  =

 

sqrt  [ 1 - [3x]^2 ]  =  sqrt [ 1 - 9x^2 ]

 

And the  cotangent  is

 

[ 3x]  / sqrt [ 1 - 9x^2 ] 

 

 

 

cool cool cool

CPhill  Oct 24, 2018
 #4
avatar+90969 
+1

cos  [ 2 arcsin ( x) ]

 

Let  arcsin (x)  be θ

 

So...using a trig identity....  we have

 

cos ( 2θ)  =    1  - 2sin^2(θ)  =  1 - 2 [ sin (θ) ]^2

 

The sin of θ  =  x / 1  =  x

 

So...

 

cos (2θ)  =  1 - 2sin^2 ( θ)  =  1 -  2 [ sin (θ)]^2  =   1  - 2 [x]^2  =  1  - 2x^2

 

 

cool cool cool   

CPhill  Oct 24, 2018

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