For what value of x is sin x = cos 19°, where 0°

Question

For what value of x is sin x = cos 19°, where 0°< x < 90°?

38°

71°

26°

19°

anonymous · Answer

@TheSmartOne  @5sauselover21

anonymous · Answer

@LeilaJudeh

freckles · Answer

think co function identity

freckles · Answer

$$\sin(90^o-x)=\cos(x)$$

anonymous · Answer

still confused......

freckles · Answer

compare cos(x) to cos(19)
what must x be?

campbell_st · Answer

look at it this way 

|dw:1428961585875:dw|

the sin of what angle given a/h

anonymous · Answer

im thinking A or C?

anonymous · Answer

nevermind figured it out! its 71 :)

thesmartone · Answer

Correct, if you did it @freckles 's way then:

$\color{blue}{\text{Originally Posted by}}$ @freckles
$$\sin(90^o-x)=\cos(x)$$
$\color{blue}{\text{End of Quote}}$

$\sf sin(90^o-x)=cos(19)$

So x=19

sin (90-19) = cos(19)

sin (71) = cos(19)

So the angle for sin that equals cos 19 is sin 71

Good job :P

freckles · Answer

|dw:1428984056958:dw|
was the point of @campbell_st 's post
his post basically was explaining the identity

freckles · Answer

I mean which is probably way better than me just stating the identity :p