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# Determining the exact angle?

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

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

Guest Feb 22, 2019
#6
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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
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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   Feb 22, 2019