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# write the trigonometric expression as an algebraic expression

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sin(arcsin x + arccos x)

how???

Guest May 7, 2017
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#1
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Note that   arcsin x   =   some angle ...  and the arccos x   will be the angle that is complementary  to  this angle

To see this....suppose  that x = 1/2....then    arcsin (1/2)  = 30°

And  arccos (1/2)     =  60°

So    arcsin x  +  arccos x   =   30° +  60°  = 90°

And   sin (90°)  =   1

CPhill  May 7, 2017
#2
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sin(arcsin x + arccos x)

i)
$$\begin{array}{lrcll} & \cos{(\varphi)} &=& x \\ \text{or}& \quad \varphi &=& \arccos{(x)}\\ \end{array}$$

ii)
$$\begin{array}{lrclcl} &\sin{(90^\circ-\varphi)} &=& \cos{(\varphi)} &=& x \\ \text{or}& \quad 90^\circ-\varphi && &=& \arcsin{(x)}\\ \end{array}$$

iii)

$$\begin{array}{rcll} && \sin\Big(\arcsin(x)+\arccos(x)\Big) \\ &=& \sin(90^\circ-\varphi+\varphi) \\ &=& \sin(90^\circ)\\ &=& 1\\ \end{array}$$

heureka  May 8, 2017
edited by heureka  May 8, 2017

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