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Angle θ is in quadrant II and sin θ \({5 \over 13}\). Determine the exact value for sin (θ + \({\pi \over 2}\)).

 

So I know this is on the triangle 5 - 12 - 13; here's what I've done:

 

sin θ + \({\pi \over 2}\) = (\({5 \over 13}\))(-\({12 \over 13}\)) + (-\({12 \over 13}\))(\({5 \over 13}\))

= - \({60 \over 169}\) - \({60 \over 169}\)

= - \({120 \over 169}\)

 

However, the answer in the textbook is - \({12 \over 13}\)What did I do wrong?

 Feb 22, 2019
 #2
avatar+17332 
+1

Sin (x + pi) =   cos x         = - 12/13    (remember ...it is in quadrant II )

 Feb 22, 2019
edited by ElectricPavlov  Feb 22, 2019
 #4
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So was my approach, using the sum and difference identity, wrong?

Guest Feb 22, 2019
 #6
avatar+17332 
+1

No....I think your APPROACH was OK....the execution of it was wrong

remember cos pi/2 = 0     sooooo    sin (x+ pi/2) = sin x cos pi/2  + cos x sin pi/2       cos pi/2 = 0   and  sin pi/2 = 1

                                                         sin ( x + pi/2) =  0  + cos x       SEE?

ElectricPavlov  Feb 22, 2019
 #5
avatar+98168 
+2

Using  sin (A + B)  = sinAcosB + sin B cosA  

 

[ cos theta = -12/13]

 

So

 

sin ( theta + pi/2 ) =   

 

sin (theta) * cos(pi/2) + sin (pi/2) * cos (theta) =

 

 (5/13) (0)  + (1) (-12/13) =

 

-12/13

 

 

cool cool cool

 Feb 22, 2019

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